Integrals

Limsup / Liminf §

Consider a sequence that oscillates between $-1$ and $1$ as $n\to \infty$. While its limit does not exist, it does still have a “notion” of two different limits! This is the idea behind limsup (limit superior) and liminf (limit inferior) - they let us make claims about the limit for sequences, which may not necessarily converge to a single value!

Let $\{x_n\}$ be bounded. Then,

1. We say the limit superior (limsup), is
   $$\underset{n\to \infty }{\mathrm{limsup}}{x}_n=\underset{n\to \infty }{\lim }\underset{k\ge n}{\sup }{x}_k$$
   Equivalent to the limit of the upper bound of $x_n$, for all $n$.
2. We say the limit inferior (liminf) is
   $$\underset{n\to \infty }{\mathrm{liminf}}{x}_n=\underset{n\to \infty }{\lim }\underset{k\ge n}{\inf }{x}_k$$
   Equivalent to the limit of the lower bound of $x_n$, for all $n$.

Note that this can be extended to unbounded sequences too! We just won’t cover them.

Note that by this definition, ${\bar{x}}_n={\sup }_{k\ge n}{x}_k$ is a decreasing sequence! And furthermore, because this sequence is bounded and decreasing, by the MCT, it converges to the infimum.

This similarly holds for the liminf.

So we can alternatively write the liminf and limsup as

$$\begin{array}{r}\underset{n\to \infty }{\mathrm{limsup}}{x}_n=\underset{n\ge 1}{\inf }(\underset{k\ge n}{\sup }{x}_k)\\ \underset{n\to \infty }{\mathrm{liminf}}{x}_n=\underset{n\ge 1}{\sup }(\underset{k\ge n}{\inf }{x}_k)\end{array}$$

Example: Computing Limsup

Suppose we have the sequence

$$x_n=\frac{(-1{)}^n}{n}$$

We find the limsup as

$${\bar{x}}_n=\underset{k\ge n}{\sup }{x}_k=\{\begin{array}{ll}(-1{)}^{n+1}/n+1& \text{n odd}\\ (-1{)}^n/n& \text{n even}\end{array}$$

Taking the limit as $n\to \infty$, this converges to 0. So, our limsup is 0.

We can similarly show that the liminf is 0.

Example: Computing Limsup (2)

Limsups are not always easy to prove! Consider

$$\underset{n\to \infty }{\lim }\sin (n)=1$$

While it seems arbitrarily, this is difficult to prove! We need to prove that $\sin (n)$ for $n\in \mathbb{N}$ is dense in $[-1,1]$.

Theorem: Addition of Limsup / Liminf

$$\begin{array}{r}\underset{n\to \infty }{\mathrm{limsup}}(x_n+y_n)\le \underset{n\to \infty }{\mathrm{limsup}}{x}_n+\underset{n\to \infty }{\mathrm{limsup}}{y}_n\\ \underset{n\to \infty }{\mathrm{liminf}}(x_n+y_n)\ge \underset{n\to \infty }{\mathrm{liminf}}{x}_n+\underset{n\to \infty }{\mathrm{liminf}}{y}_n\end{array}$$

Proof (Limsup)

By the upper bound nature of supremums,

$$x_n+y_n\le \underset{k\ge n}{\sup }{x}_k+\underset{k\ge n}{\sup }{y}_k$$

And furthermore, as the supremum of $x_n+y_n$ is the least upper bound,

$$\underset{k\ge n}{\sup }(x_k+y_k)\le \underset{k\ge n}{\sup }{x}_k+\underset{k\ge n}{\sup }{y}_k$$

We apply the limit on both sides to obtain our theorem.

The limit set of $\{x_n{\}}_{n=1}^{\infty }$ is

$$\{x\in \mathbb{R}|\exists x_{n_j}\to x\}$$

As $j\to \infty$. In other words, its the set of all limits that $\{x_n\}$’s subsequences converge to.

Example: Limit Sets

$$x_n=\{\begin{array}{ll}\frac{5}{n+1}& \text{n odd}\\ \frac{(-1{)}^n+n^2}{n^2}& \text{n even}\end{array}$$

Then, our limit set is $S=\{0,1\}$.

Interestingly enough, in the above example, our liminf is 0, and our limsup is 1! This is no coincidence - we can use limit sets to compute our liminf and limsup!

Theorem: Limit Sets and Liminf / Limsup

Let $\{x_n\}$ be a bounded sequence, and let $S$ be the limit set.

Then,

$$\begin{array}{r}\underset{n\to \infty }{\mathrm{limsup}}{x}_n=\sup (S)\\ \underset{n\to \infty }{\mathrm{liminf}}{x}_n=\inf (S)\end{array}$$

We omit the proof.

Theorem: Convergence and Limsup / Liminf

$$\mathrm{limsup}{x}_n=\mathrm{liminf}{x}_n⟺\lim x_n=L\text{ exists}$$

In which case, all 3 limits are equal to $L$.

Proof (Forward Direction)

Proof ($\to$) §

Assume $\mathrm{liminf}{x}_n=\mathrm{limsup}{x}_n=L$. Then, for any $n$,

$$\underset{k\ge n}{\inf }{x}_k\le x_n\le \underset{k\ge n}{\sup }{x}_k$$

Taking the limit of both sides as $n\to \infty$, we get

$$L\le \underset{n\to \infty }{\lim }{x}_n\le L$$

Forcing our limit to be equal to $L$.

Integration §

Darboux Integrals §

Suppose we have a function $f:[a,b]\to \mathbb{R}$. In this section, we formally define what an integral is.

$${\int }_a^{b}f(x)dx$$

Definition of an Integral

Recall that in earlier calculus classes, we define an integral as the sum of progressionally smaller rectangles under the curve, known as a Riemann Sum.

$${\int }_a^{b}f(x)dx=\underset{n\to \infty }{\lim }\sum _{i=1}^{n}f(x_i)\Delta x_i$$

Consider the closed interval $[a,b]$, and partition it using $n+1$ points $x_0,x_1,x_2,\dots x_n\in [a,b]$, where

$$x_0=a{x}_i<x_{i+1},\forall i{x}_n=b$$

Let $f$ be bounded, and consider the interval between any two partition points $[x_{i-1},x_i]$. By assumption, we can define the quantities $m_i$, the infimum of $f$ along this interval, and $M_i$, the supremum of $f$ along this interval.

$$m_i=\underset{x\in [x_{i-1},x_i]}{\inf }f(x)M_i=\underset{x\in [x_{i-1},x_i]}{\sup }f(x)$$

And use this to define the lower ($L$) and upper ($U$) Darboux sums of $f$,

$$L(f,P)=\sum _{i=1}^n{m}_i\Delta x_{i}U(f,P)=\sum _{i=1}^n{M}_i\Delta x_i$$

Where $\Delta x_i=x_i-x_{i-1}$.

Note that as $\Delta x_i$ is the width of our interval, we are defining rectangles under $f$ (whose heights are either the infimum of supremum along the interval)! This is very similar to Riemann Sums.

Theorem: Darboux Sums

Suppose $m\le f(x)\le M$ for all $x\in [a,b]$. Then,

$$m(b-a)\le L(f,P)\le U(f,P)\le M(b-a)$$

Proof

$P=\{x_0,\dots x_n\}$ be a partition of $[a,b]$, where $x_i$ is a subinterval, and let $m_i$, $M_i$ be the infimum and supremums along this subinterval. Then,

$$\begin{array}{r}m\Delta x_i\le f(x)\Delta x_i\le M\Delta x_i\\ \Delta x_i\le m_i\Delta x_i\le M_i\Delta x_i\le M\Delta x_i\end{array}$$

Summing this over all of the subintervals, we get

$$m(b-a)\le L(f,P)\le U(f,P)\le M(b-a)$$

Now let $P$ be a partition of $[a,b]$. Then, we say $P^{\ast }$ is a refinement of $P$ if $P^{\ast }$ contains all points of $P$, and possibly others.

A refinement is essentially just a partition with more points inside it.

Example: Refinement

$$\left\{0,\frac{1}{3},\frac{1}{2},1\right\}$$

The below is an example of a refinement of this partition

$$\left\{0,\frac{1}{3},\frac{1}{2},\frac{3}{4},1\right\}$$

Refinements are important, as they let us define finer and finer sums for $L$ and $U$!

Theorem: Refinements and Darboux Sums

Let $P$ be a partition of $[a,b]$, and $P^{\ast }$ refine $P$. Then,

$$\begin{array}{r}L(f,P)\le L(f,P^{\ast })\\ U(f,P)\ge U(f,P^{\ast })\end{array}$$

This should intuitively make sense! If we more finely divide up $[a,b]$, then the minimum of $f$ along our new subintervals can only increase, and the maximum of $f$ along these subintervals can only decrease!

Theorem: Partitions and Darboux Sums

For any partitions $P_1$ and $P_2$ of $[a,b]$, lower sums are always less than upper sums.

$$L(f,P_1)\le U(f,P_2)$$

Proof

Define the common refinement of $P_1$ and $P_2$, by combining them as $P^{\ast }=P_1\cup P_2$. Then, we can apply our previous theorem to obtain

$$L(f,P_1)\le L(f,P^{\ast })\le U(f,P^{\ast })\le U(f,P_2)$$

We can use these theorems to define what an integral is!

Let $P$ be an arbitrary partition of $[a,b]$. Then, we define the lower / upper (Darboux) integral as

$$\underline{{\int }_a^b}f=\underset{P}{\sup }\{L(f,P)\}\overline{{\int }_a^b}f=\underset{P}{\inf }\{U(f,P)\}$$

Or in other words, the maximum lower sum, and the minimum upper sum over all possible partitions $P$.

Note that the $P$ underneath the supremums and infimums is notation! We’re essentially applying a “supremum / infimum” over all possible $L$ and $U$, where $P$ is arbitrary (like the $dx$ in an integral).

If these two integrals are equal, then we say $f$ is integrable (in terms of Darboux) and we write

$${\int }_a^{b}f=\underline{{\int }_a^b}f=\overline{{\int }_a^b}f$$

for the commmon value.

Theorem: Darboux Sums and Integrals

Let $f:[a,b]\to \mathbb{R}$ be a function. Then,

$$L(f,P)\le \underline{{\int }_a^b}f\le \overline{{\int }_a^b}f\le U(f,P)$$

Proof

The first and last inequalities naturally hold by definition of infimums and supremums.

We apply our theorem above. For any two arbitrary partitions $P_1,P_2$,

$$L(f,P_1)\le U(f,P_2)$$

Because $P_1$ and $P_2$ are arbitrary, we can apply our supremum on $P_1$ to obtain

$$\underset{P_1}{\sup }L(f,P_1)\le U(f,P_2)$$

We know that the supremum exists, as $L(f,P_1)$ is bounded above by $U$.

By the same idea, we apply our infimum on $P_2$ to obtain

$$\underset{P_1}{\sup }L(f,P_1)\le \underset{P_2}{\inf }U(f,P_2)$$

Which translates to our integrals by definition!

Example: Computing Darboux Integrals

Consider the constant function over interval $[a,b]$, $f(x)=c$.

We find our integral.

$$\begin{array}{rl}& L(f,P)=\sum _{i=1}^n{m}_i\Delta x_i=\sum _{i=1}^{n}c\Delta x_i=c(b-a)\\ & U(f,P)=\sum _{i=1}^n{M}_i\Delta x_i=\sum _{i=1}^{n}c\Delta x_i=c(b-a)\\ & \underset{P}{\sup }L(f,P)=c(b-a)=\underset{P}{\inf }U(f,P)\\ & {\int }_a^{b}f=c(b-a)\end{array}$$

Example: Computing Darboux Integrals

Consider the Dirichlet function

$$f(x)=\{\begin{array}{ll}0& x\in \mathbb{Q}\\ 1& x\notin \mathbb{Q}\end{array}$$

Is $f$ integrable along $[0,1]$?

Because the rationals / irrationals are dense, any subinterval will always contain a rational and an irrational, so $f$ will always be both 0 and 1 at some point in the interval.

$$\begin{array}{rl}& L(f,P)=\sum _{i=1}{m}_i\Delta x_i=\sum _{i=1}0\Delta x_i=0\\ & \underset{P}{\sup }L(f,P)=0\\ & U(f,P)=\sum _{i=1}{M}_i\Delta x_i=\sum _{i=1}1\Delta x_i=(1-0)\\ & \underset{P}{\inf }U(f,P)=1\\ & \underset{P}{\sup }L(f,P)\ne \underset{P}{\inf }U(f,P)\end{array}$$

As these are not equal, our function is not integrable along $[0,1]$ in the sense of the Riemann Darboux.

Archimedes-Riemann Theorem §

Computing these integrals can be annoying, especially as we need to compute infimums and supremums! The below theorem can help simplify our integral computations in specific cases.

Theorem: Archimedes-Riemann Theorem

Let $f:[a,b]\to \mathbb{R}$ be bounded. Then, $f$ is integrable if and only if there exists a sequence of partitions $\{P_n\}$ such that

$$\underset{n\to \infty }{\lim }(U(f,P_n)-L(f,P_n))=0$$

In this case, we say that $\{P_n\}$ is archimedian.

Furthermore, in the case that it is integrable,

$${\int }_a^{b}f=\underset{n\to \infty }{\lim }U(f,P_n)=\underset{n\to \infty }{\lim }L(f,P_n)$$

Proof

Proof ($\leftarrow$) §

Suppose there exists a sequence of partitions $\{P_n\}$ such that

$$\{U(f,P_n)-L(f,P_n)\}\to 0$$

We know that

$$\begin{array}{r}\overline{{\int }_a^b}f\le U(f,P_n)\\ \underline{{\int }_a^b}f\ge L(f,P_n)\end{array}$$

So we use these to obtain

$$0\le \overline{{\int }_a^b}-\underline{{\int }_a^b}f\le U(f,P_n)-L(f,P_n)\to 0$$

So by squeeze theorem,

$$\overline{{\int }_a^b}f=\underline{{\int }_a^b}f$$

Proof ($\to$) §

Suppose

$$\overline{{\int }_a^b}f=\underline{{\int }_a^b}f={\int }_a^{b}f$$

By definition,

$$\overline{{\int }_a^b}f=\underset{P}{\inf }U(f,P)\underline{{\int }_a^b}f=\underset{P}{\sup }L(f,P)$$

So by the epsilon-definition of supremums, we know that $\forall ϵ>0$, there exists a $L(f,P)$, $U(f,P)$ such that

$$L(f,P)<\underset{P}{\inf }U(f,P)+ϵ\underset{P}{\sup }L(f,P)-ϵ<U(f,P)$$

We choose $ϵ=1/n$, and for all $n\in \mathbb{N}$ choose these $L$’s and $U$’s to find a sequence. Note that these sequences converge by squeeze theorem.

$$\begin{array}{r}\underset{P}{\inf }U(f,P)<L(f,P_n)<\underset{P}{\inf }U(f,P)+\frac{1}{n}\\ \underset{P}{\sup }L(f,P)-\frac{1}{n}<U(f,P_n^{\prime })<\underset{P}{\sup }L(f,P)\end{array}$$

Our partitions are not guaranteed to be the same, so we can take the union of them, $P_n^{\prime }\cup P_n=Q_n$

$$\begin{array}{r}\underset{P}{\inf }U(f,P)<L(f,Q_n)\le L(f,P_n)<\underset{P}{\inf }U(f,P)+\frac{1}{n}\\ \underset{P}{\sup }L(f,P)-\frac{1}{n}<U(f,P_n^{\prime })\le U(f,Q_n)<\underset{P}{\sup }L(f,P)\end{array}$$

Finally,

$$\begin{array}{rl}0& \le U(f,Q_n)-L(f,Q_n)\\ & <{\int }_a^{b}f+\frac{1}{n}+\frac{1}{n}-{\int }_a^{b}f\\ & =\frac{2}{n}\to 0\end{array}$$

We apply squeeze theorem to obtain convergence.

We now show that

$${\int }_a^{b}f=\underset{n\to \infty }{\lim }U(f,P_n)$$

To do this, we use the property of infimums that

$$\begin{array}{r}\overline{{\int }_a^b}f\le U(f,P_n)\\ 0\le U(f,P_n)-\overline{{\int }_a^b}f\le U(f,P_n)-L(f,P_n)\end{array}$$

As $n\to 0$, we apply squeeze theorem.

$$U(f,P_n)-\overline{{\int }_a^b}=0⟹\overline{{\int }_a^b}=U(f,P_n)$$

We have similar argument for the limit of the lower sum!

This theorem is really important, and can be used to prove a lot! In fact, many of the below properties of integrals can be proven using the Archimedes-Riemann Theorem. See below.

Properties of Integrals §

Below, we discuss various useful properties of integrals.

Monotonicity and Integrals §

Theorem: Monotonicity and Integrability

Any monotone function $f:[a,b]\to \mathbb{R}$ is integrable.

The monotonicity on $f$ and its restricted domain automatically implies boundedness!

Proof

Let $P_n=\{x_0,x_1,\dots x_n\}$ where $\Delta x_i=x_i-x_{i-1}=(b-a)/n$ is constant.

We show that $\{P_n\}$ is Archimedian, to apply the previous theorem.

$$\begin{array}{rl}0& \le U(f,P_n)\Delta x-L(f,P_n)\\ & =\sum _{i=1}^n{M}_i\Delta x-\sum _{i=1}^n{m}_i\Delta x\\ & =\frac{b-a}{n}\left(\sum _{i=1}^{n}f(x_i)-\sum _{i=1}^{n}f(x_{i-1})\right)\\ & =\frac{b-a}{n}(f(x_n)-f(x_0))\\ & =\frac{b-a}{n}(f(b)-f(a))\to 0\end{array}$$

Thus, we apply our above theorem to obtain integrability.

We can use a similar idea to show that any step function is integrable, though this is much more difficult.

Lemma

Let $\{P_n\}$ be Archimedian for $f:[a,b]\to \mathbb{R}$. Then any refinement $\{P_n^{\ast }\}$ is also Archimedian.

Proof

By property of refinements, $U$ will get smaller, and $L$ will get larger, so

$$\begin{array}{rl}0& \le U(f,P_n^{\ast })-L(f,P_n^{\ast })\\ & \le U(f,P_n)-L(f,P_n)\to 0\end{array}$$

Theorem: Monotonicity

Let $f(x)\le g(x)$ for all $x\in [a,b]$. Then,

$${\int }_a^{b}f\le {\int }_a^{b}g$$

Proof

Let $\{P_n\}$ be Archimedian for $f$, and $\{P_n^{\prime }\}$ for $g$. Let $Q_n=P_n\cup P_n^{\prime }$, which is still Archimedian for $f$ and $g$!

With a common refinement, we have $U(f,Q_n)\le U(g,Q_n)$. Then, as $n\to \infty$, by the Archimedian Riemann theorem, we are done.

Additivity, Linearity, Absolute Value of Integrals §

Theorem: Additivity of Integrals

Let $f:[a,b]\to \mathbb{R}$ be an integrable function. Then, $\forall c\in (a,b)$,

$${\int }_a^{b}f={\int }_a^{c}f+{\int }_c^{b}f$$

Proof

Suppose there exists an Archimedian Sequence $\{P_n\}$ for $f$ on $[a,b]$. Without loss of generality, suppose $\{P_n\}$ contains $c$ (since any refinement is also Archimedian).

Split $\{P_n\}$ into mutually exclusive partitions $P_n^{\prime }$ and $P_n^{\prime \prime }$ based on $c$. We want to show that both subsequences are Archimedian.

We have

$$\begin{array}{rl}& U(f,P_n)-L(f,P_n)\to 0\\ & (U(f,P_n^{\prime })+U(f,P_n^{\prime \prime }))-(L(f,P_n^{\prime })+L(f,P_n^{\prime \prime }))\to 0\\ & (U(f,P_n^{\prime })-L(f,P_n^{\prime }))+(U(f,P_n^{\prime \prime })-L(f,P_n^{\prime \prime }))\to 0\end{array}$$

Because $U(f,P_n^{\prime })-L(f,P_n^{\prime })$ and $U(f,P_n^{\prime \prime })-L(f,P_n^{\prime \prime })$ are both non-negative for all $n$, for this sequence to converge it must be true that they both converge to 0.

Theorem: Linearity of Integrals

Let $\alpha ,\beta \in \mathbb{R}$, and $f,g:[a,b]\to \mathbb{R}$ be integrable. Then,

$${\int }_a^b(\alpha f+\beta g)=\alpha {\int }_a^{b}f+\beta {\int }_a^{b}g$$

Proof (Half)

We show the first part of the proof.

$${\int }_a^{b}f+g={\int }_a^{b}f+{\int }_a^{b}g$$

Let $P_n$ be an Archimedian sequence for $f$ and $g$ on $[a,b]$, which is possible by common refinement.

We know that

$$\begin{array}{r}f(x)+g(x)\le \sup f(x)+\sup g(x)\\ \sup (f(x)+g(x))\le \sup f(x)+\sup g(x)\\ U(f+g,P_n)\le U(f,P_n)+U(g,P_n)\end{array}$$

Similarly, we can obtain

$$L(f+g,P_n)\ge L(f,P_n)+L(g,P_n)$$

Giving us

$$\begin{array}{rlr}& L(f,P_n)+L(g,P_n)\le L(f+g,P_n)\le U(f+g,P_n)\le U(f,P_n)+U(g,P_n)& \\ & {\int }_a^{b}f+{\int }_a^{b}f\le {\int }_a^{b}f+g{\int }_a^{b}f+g\le {\int }_a^{b}f+{\int }_a^{b}g& n\to \infty \\ & {\int }_a^{b}f+{\int }_a^{b}g={\int }_a^{b}f+g& \end{array}$$

Theorem: Absolute Value and Integrals

Let $f$ and $|f|$ be integrable on $[a,b]$. Then,

$$\left|{\int }_a^{b}f\right|\le {\int }_a^b|f|$$

Proof

We know that

$$-|f(x)|\le f(x)\le |f(x)|$$

By monotonicity,

$$-{\int }_a^b|f|\le {\int }_a^{b}f\le {\int }_a^b|f|$$

But this is essentially an absolute value!

$$\left|{\int }_a^{b}f\right|\le {\int }_a^b|f|$$

Continuity and Integrals §

The follow theorems guarantee integrability for continuous functions.

Theorem: Continuity and Integrability

Let $f$ be a continuous function on compact interval $[a,b]$. Then, its integral along this interval exists.

Proof

Since we have a continuous function on a compact interval, we know $f$ is uniformly continuous. By definition, $\forall ϵ>0,\exists \delta >0$ such that $\forall x,y\in [a,b]$,

$$|x-y|<\delta \to |f(x)-f(y)|<ϵ$$

For later convenience, take the constant $\frac{ϵ}{b-a}$.

Let $\{P_n\}=\{x_0,\dots ,x_n\}$ be a uniform sequence of partitions for $[a,b]$, where $x_0=a,x_n=b$. As the partition is uniform, we have that $\Delta x_i=\frac{b-a}{n}$. We wish to show that $\{P_n\}$ is Archimedian.

We know that by a theorem,

$$\begin{array}{rl}0& \le U(f,P_n)-L(f,P_n)\\ & \le \sum _{i=1}^n\left(\underset{x\in [x_{i-1},x_i]}{\sup }f\right)\Delta x_i-\sum _{i=1}^n\left(\underset{x\in [x_{i-1},x_i]}{\inf }f\right)\Delta x_i\end{array}$$

By the Extreme Value theorem, we know that for some $u_i,v_i\in [x_{i-1},x_i]$,

$$\underset{x\in [x_{i-1},x_i]}{\sup }f=f(u_i)\underset{x\in [x_{i-1},x_i]}{\inf }f=f(v_i)$$

This gives us

$$\begin{array}{rl}0& \le \sum _{i=1}^n\left(\underset{x\in [x_{i-1},x_i]}{\sup }f\right)\Delta x_i-\sum _{i=1}^n\left(\underset{x\in [x_{i-1},x_i]}{\inf }f\right)\Delta x_i\\ & \le \sum _{i=1}^n(f(u_i)-f(v_i))\Delta x_i\\ & \le \underset{1\le i\le n}{\max }(f(u_i)-f(v_i))\sum _{i=1}^n\Delta x_i\\ & \le f(u_{i_0})-f(v_{i_0})(b-a)\end{array}$$

Where $u_{i_0},v_{i_0}$ is the greatest difference between the maximum and minimum.

If $|u_{i_0}-v_{i_0}|<\delta$, we can apply uniform continuity, we have

$$f(u_{i_0})-f(v_{i_0})(b-a)<\frac{ϵ}{b-a}(b-a)<ϵ$$

And since this holds for all $ϵ$, we can let $ϵ\to 0$ and apply squeeze theorem, to have

$$\{U(f,P_n)-L(f,P_n)\}\to 0$$

To guarantee the $|u_{i_0}-v_{i_0}|<\delta$ condition, we choose $N$ large enough such that $\Delta x_i<\delta$, so that $\Delta x_i<\delta$ for all $n\ge N$.

Theorem: Continuity and Integrability

Let $f$ be continuous on open $(a,b)$ and bounded on $[a,b]$. Then, ${\int }_a^{b}f$ exists and does NOT depend on $f(a)$ or $f(b)$.

Proof (Sketch)

Create an interval $I_n=[a+\frac{1}{n},b-\frac{1}{n}]$. For each $I_n$, choose a partition $\{P_n^{\ast }\}$ satisfying

$$0\le U(f,P_n^{\ast })-L(f,P_n^{\ast })\le \frac{1}{n}$$

Whose existence is guaranteed by our previous proof.

Let $P_n=P_n^{\ast }\cup \{a,b\}$. We show that $\{P_n\}$ is Archimedian for $[a,b]$ to show that ${\int }_a^{b}f$ exists, and then show that

$$U(f,P_n^{\ast })\to {\int }_a^{b}f$$

Since $P_n^{\ast }$ doesn’t depend on $a$ or $b$, it must be true that the integral also doesn’t depend on $f(a)$ or $f(b)$!

Example: Continuity and Integrability

$$f=\{\begin{array}{ll}\sin (1/x)& x\in (0,1)\\ 0& x=0\end{array}$$

As $f$ is continuous on $(0,1)$, and bounded on $[0,1]$, its integral ${\int }_0^{1}f$ exists.

Fundmental Theorems of Calculus §

Theorem: First Fundamental Theorem of Calculus

Let $F$ be a continuous function on $[a,b]$, and $F^{\prime }$ be bounded and continuous on $[a,b]$.

Then,

$${\int }_a^b{F}^{\prime }(x)dx=F(b)-F(a)$$

Proof

Since $F^{\prime }$ is continuous and bounded, our integral ${\int }_a^b{F}^{\prime }$ exists by a theorem.

Let $P=\{x_0,\dots ,x_n\}$ be a partition. Because $F$ is continuous, we apply Mean Value Theorem on each subinterval $[x_{i-1},x_i]$, we know that $\exists c_i$ such that

$$F(x_i)-F(x_{i-1})=F^{\prime }(c_i)\Delta x_i$$

Since $F^{\prime }$ is bounded, we know that

$$\begin{array}{rl}\underset{[x_{i-1},x_i]}{\inf }{F}^{\prime }(x)\le & F^{\prime }(c_i)\le \underset{[x_{i-1},x_i]}{\sup }{F}^{\prime }(x)\\ m_i\le & F^{\prime }(c_i)\le M_i\end{array}$$

So $m_i\Delta x_i\le F(x_i)-F(x_{i-1})\le M_i\Delta x_i$. We take the sum over all the subintervals to get

$$L(F^{\prime },P)\le F(b)-F(a)\le U(F^{\prime },P)$$

As this holds for all lower sums, we have

$$\begin{array}{r}\underset{P}{\sup }L(F^{\prime },P)\le F(b)-F(a)\le \underset{P}{\inf }U(F^{\prime },P)\\ \underline{{\int }_a^b}{F}^{\prime }\le F(b)-F(a)\le \overline{{\int }_a^b}{F}^{\prime }\end{array}$$

And as we assume integrability, this gives us

$${\int }_a^b{F}^{\prime }\le F(b)-F(a)\le {\int }_a^b{F}^{\prime }⟹{\int }_a^b{F}^{\prime }=F(b)-F(a)$$

We need the below theorems to prove the second fundamental theorem.

Lemma

If $a>b$, then

$${\int }_a^{b}f=-{\int }_b^{a}f$$

Theorem: Integral Mean Value Theorem

If $f$ is a continuous function on $[a,b]$, then $\exists c\in (a,b)$ such that

$${\int }_a^{b}f=f(c)(b-a)$$

Proof

By EVT, $f(x_{\text{min}})\le f(x)\le f(x_{\text{max}})$. We integrate this to obtain

$$\begin{array}{r}f(x_{\text{min}})(b-a)\le {\int }_a^{b}f\le f(x_{\text{max}})(b-a)\\ f(x_{\text{min}})\le \frac{1}{(b-a)}{\int }_a^{b}f\le f(x_{\text{max}})\end{array}$$

Now, for all points between the minimum and maximum, we can find a $c$ by intermediate value theorem equal to $f$ at that point - finishing our proof.

Theorem: Second Fundamental Theorem of Calculus

If $f$ is a continuous function on $[a,b]$, then for any $x\in (a,b)$,

$$\frac{d}{dx}{\int }_a^{x}f(t)dt=f(x)$$

Proof

Let $\{x_n\}\to x_0$, $\{x_n\}\in [a,b]/\{x_0\}$. By definition of derivative,

$$\begin{array}{rl}\frac{{\int }_a^{x_n}f-{\int }_a^{x_0}f}{x_n-x_0}& =\frac{{\int }_{x_0}^{x_n}f}{x_n-x_0}\\ & =\frac{1}{x_n-x_0}f(C_n){\int }_{x_0}^{x_n}1dx\\ & =f(C_n)\end{array}$$

By the Integral Mean Value Theorem where $C_n$ is between $x_0$ and $x_n$. We evaluate this and take the limit to obtain $f(x_0)$.

Extra Integral Properties §

Theorem: Generalized Integral MVT

Let $g$ be integrable and $g(x)>0$, and let $f:[a,b]\to \mathbb{R}$ be continuous. Then, $\exists x_0\in (a,b)$ such that

$${\int }_a^{b}fg=f(x_0){\int }_a^{b}g$$

Note that if $g=1$, we have the Integral MVT!

Proof

By assumption, we know that $f:[a,b]\to \mathbb{R}$. Thus, by the Extreme Value Theorem, $f$ has a minimum and maximum at $x_{\text{min}},x_{\text{max}}$ (respectively).

By property of minimums and maximums, we know that $\forall x\in [a,b]$,

$$\begin{array}{rlr}f(x_{\text{min}})\le & f(x)\le f(x_{\text{max}})& \text{Mins and Maxes}\\ f(x_{\text{min}})g(x)\le & f(x)g(x)\le f(x_{\text{max}})g(x)& g(x)\ge 0\\ {\int }_a^{b}f(x_{\text{min}})g(x)\le {\int }_a^b& f(x)g(x)\le {\int }_a^{b}f(x_{\text{max}})g(x)& \text{Monotonicity}\\ f(x_{\text{min}}){\int }_a^{b}g(x)\le {\int }_a^b& f(x)g(x)\le f(x_{\text{max}}){\int }_a^{b}g(x)& \text{Constants}\\ f(x_{\text{min}})\le & \frac{{\int }_a^{b}f(x)g(x)}{{\int }_a^{b}g(x)}\le f(x_{\text{max}})& \end{array}$$

Since this value is bounded by the minimum and maximum of $f$,by the Intermediate Value Theorem, we know that there exists a $x_0\in (a,b)$ such that

$$\begin{gathered} f(x_0)=\frac{{\int }_a^{b}f(x)g(x)}{{\int }_a^{b}g(x)} \\ ⟹{\int }_a^{b}f(x)g(x)=f(x_0){\int }_a^{b}g(x) \end{gathered}$$

Theorem: Continuity of Integration

Define $F(x)={\int }_a^{x}f(t)dt$ where $f$ is integrable. Then $F(x)$ is a continuous function, and in fact is uniformly continuous!

Proof (Sketch)

Let $F(x)={\int }_a^{x}f(t)dt$. Without loss of generality, say $x>y$. We have

$$\begin{array}{rlr}|F(x)-F(y)|& =\left|{\int }_a^{x}f-{\int }_a^y\right|=\left|{\int }_y^{x}f\right|& \\ & \le {\int }_y^x|f|& \text{Property of Absolute Values}\\ & \le \underset{a\le x\le b}{\max }|f|\cdot (x-y)& \end{array}$$

Choose $\delta =\frac{1}{{\max }_{a\le x\le b}|f|\cdot (x-y)}$ to obtain uniform continuity.

Theorem: Fundamental Theorem of Calculus + Chain Rule

$$\frac{d}{dx}{\int }_{a(x)}^{b(x)}f(t)dt=f(b(x))\frac{db}{dx}-f(a(x))\frac{da}{dx}$$

We can easily prove this by splitting our integral along a middle $c$ value, and applying the second fundamental theorem.

Integration by Parts §

Theorem: Integration by Parts

If $f$ and $g$ are continuous on $[a,b]$, and $f^{\prime }$ and $g^{\prime }$ are bounded and continuous on $(a,b)$. Then

$$\int f{g}^{\prime }=fg-\int g{f}^{\prime }$$

Additionally,

$${\int }_a^{b}f{g}^{\prime }=fg{|}_{x=a}^{x=b}-{\int }_a^b{f}^{\prime }g$$

What we are essentially doing with integration by parts is “moving” the derivative on $g$ onto $f$!

Proof

$$\begin{array}{r}(fg{)}^{\prime }=f{g}^{\prime }+g{f}^{\prime }\\ fg=\int f{g}^{\prime }+\int g{f}^{\prime }\end{array}$$

Example: Integration by Parts

$$\begin{array}{rl}\int x{e}^x& =\int x\frac{d}{dx}{e}^{x}dx\\ & =x{e}^x-\int e^x\frac{d}{dx}xdx\\ & =x{e}^x-e^x+C\end{array}$$

Theorem: u-Substitution

Let $f:[a,b]\to \mathbb{R}$ be continuous, $g:[c,d]\to \mathbb{R}$ be continuous, $g^{\prime }$ continuous and bounded on $(c,d)$, and $\text{range}(g)\subset (a,b)$. Then,

$${\int }_c^{d}f(g(t))g^{\prime }(t)dt={\int }_{g(c)}^{g(d)}f(u)du$$

Proof

Let $H(x)={\int }_c^{x}f(g(t))g^{\prime }(t)-{\int }_{g(c)}^{g(x)}f(u)du$. We show it is equal to the 0 function to obtain our result.

First, note that $H(c)=0$. By the second fundamental theorem of calculus,

$$\frac{d}{dx}H(x)=f(g(x))g^{\prime }(x)-f(g(x))g^{\prime }(x)=0$$

And as the derivative is also 0, this forces $H(x)=0$ for all $x$.

Thus, we have

$${\int }_c^{x}f(g(t))g^{\prime }(t)-{\int }_{g(c)}^{g(x)}f(u)du=0$$

Let $x=d$ and rearrange to get our result.