ANA1_L27

Recall

For any bounded $f:[a,b]\to \mathbb{R}$, a partition $P$ of $[a,b]$, a refinement $P^{\ast }$ of $P$, and monotone increasing function $\alpha :[a,b]\to \mathbb{R}$, we have

$$\begin{array}{rl}m(\alpha (b)-\alpha (a))& \le \\ & \\ L(P,f,\alpha )\le L(P^{\ast },f,\alpha )& \le \\ & \\ \underline{{\int }_a^b}fd\alpha \le \overline{{\int }_a^b}fd\alpha & \le \\ & \\ U(P^{\ast },f,\alpha )\le U(P,f,\alpha )& \le \\ & \\ M(\alpha (b)-\alpha (a)).& \end{array}$$

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Which functions are integrable?

Example 1(Example of a non integrable function).

Consider the function $f$ on $\mathbb{R}$ which is $1$ at rational inputs and $0$ at irrational inputs. For any partition, each $M_i=1$ and each $m_i=0$. Thus, the upper Riemann integral will be $(b-a)$, while the lower Riemann integral will be $0$. Thus, this function is not integrable over any interval.

Continuous functions

Rudin, 6.8

Theorem 2(Theorem).

If $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ then $f\in \mathcal{R}(\alpha )$ on $[a,b]$.

Proof
Note that since $[a,b]$ is compact, $f$ is uniformly continuous on $[a,b]$. We will use the the criterion for integrability to show $f\in \mathcal{R}$. Let $ϵ>0$. We can find $\delta >0$ such that whenever $|s-t|<\delta$, $|f(s)-f(t)|<ϵ$. Choose $P$ such that each $\Delta x_i<\delta$. Thus, in $[x_{i-1},x_i]$, $M_i-m_i<ϵ$. Therefore,

$$\begin{array}{rl}U(P,f,\alpha )-L(P,f,\alpha )=& \sum _{i=1}^n(M_i-m_i)\Delta {\alpha }_i<ϵ(\alpha (b)-\alpha (a)).\end{array}$$

Thus, $f\in \mathcal{R}(\alpha )$. ❏

Note that $M_i-m_i$ cannot be equal to $ϵ$, since $f$ actually attains these extrema in $[x_{i-1},x_i]$ due to the extreme value theorem (this doesn’t impact the proof in any way; just an observation).

Monotonic functions

Theorem 3(Theorem).

If $f:[a,b]\to \mathbb{R}$ is monotone on $[a,b]$ then $f\in \mathcal{R}$.

Proof
Let $f$ be monotone increasing. For any $P$, we have

$$\begin{array}{rl}U(P,f)=& f(x_1)\Delta x_1+f(x_2)\Delta x_2+\cdots +f(x_n)\Delta x_n\\ L(P,f)=& f(x_0)\Delta x_1+f(x_1)\Delta x_2+\cdots +f(x_{n-1})\Delta x_n\end{array}$$

Consider $P$ such that all $\Delta x_i$ are equal to $\frac{b-a}{n}$ for some $n\in \mathbb{N}$. Then,

$$U(P,f)-L(P,f)=(f(x_n)-f(x_0))\frac{b-a}{n}.$$

$n$ can be made arbitrarily large to meet any $ϵ$ challenge. ❏

Note that it does not matter whether $f$ is continuous or not.

Rudin, 6.9

Theorem 4(Theorem).

If $f:[a,b]\to \mathbb{R}$ is monotone on $[a,b]$ and $\alpha$ is continuous on $[a,b]$ then $f\in \mathcal{R}(\alpha )$.

Proof
Let $f$ be monotone increasing. For any $P$, we have

$$\begin{array}{rl}U(P,f,\alpha )=& f(x_1)\Delta {\alpha }_1+f(x_2)\Delta {\alpha }_2+\cdots +f(x_n)\Delta {\alpha }_n\\ L(P,f,\alpha )=& f(x_0)\Delta {\alpha }_1+f(x_1)\Delta {\alpha }_2+\cdots +f(x_{n-1})\Delta {\alpha }_n\end{array}$$

Now, if we are to pull the same trick as we did in the previous proof, we need $\alpha$ to play nice, i.e,

$$\forall n\in \mathbb{N},\forall 0\le i\le n,\exists x_i\in [a,b]\text{ such that }\alpha (x_i)=\alpha (a)+\frac{i}{n}(\alpha (b)-\alpha (a)).$$

This is similar to the conclusion of the intermediate value theorem, but there is a key difference in that we only require $x_i$ to exist for a specific subset of rational numbers, and not all numbers in $[\alpha (a),\alpha (b)]$. However, the intermediate value theorem holding would suffice. We can ensure this by requiring $\alpha$ to be continuous (If a monotone function has IVP, then it must be continuous), as done in the hypothesis of this theorem. So, we can proceed to write

$$U(P,f,\alpha )-L(P,f,\alpha )=(f(x_n)-f(x_0))\frac{\alpha (b)-\alpha (a)}{n},$$

and make $n$ as small as we like to meet any $ϵ$ challenge. ❏

Note that $\alpha$ being a derivative also works - derivatives have the IVP.

Continuous functions with finitely many transgressions

Rudin, 6.10

Theorem 5(Theorem).

Suppose $f$ is continuous on $[a,b]$ except at finitely many points, and $f$ is bounded on $[a,b]$. Then, $f\in \mathcal{R}$ on $[a,b]$.

Proof
It suffices to consider the case when $f$ is discontinuous at only one point in $[a,b]$, say $p$. Let $ϵ>0$. Now, let $k\in (p-ϵ,p)$ and $k^{\prime }\in (p,p+ϵ)$. $f$ is uniformly continuous on $[a,k]\cup [k^{\prime },b]$, so we can choose $\delta$ such that $s,t\in [a,k]\cup [k^{\prime },b],|s-t|<\delta ⟹|f(s)-f(t)|<ϵ$. Now, choose a partition $P$ such that $x_{j-1}=k$ and $x_j=k^{\prime }$ for some $j$, and $\Delta x_i<\delta$ for all $i\ne j$. Let $M=\sup |f(x)|$. Note that $M_j-m_j\le 2M$. Then,

$$\begin{array}{rl}U(P,f)-L(P,f)=& \sum _{i=1}^{j-1}(M_i-m_i)(x_i-x_{i-1})\\ & +(M_j-m_j)(x_j-x_{j-1})\\ & +\sum _{i=j+1}^n(M_i-m_i)(x_i-x_{i-1})\\ \le & ϵ(x_{j-1}-a)+2M(2ϵ)+ϵ(b-x_j)\\ =& ϵ(-(x_j-x_{j-1})+b-a+4M)\\ <& ϵ(b-a+4M)\end{array}$$

❏

Theorem 6(Theorem).

Suppose $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ except at finitely many points, and $\alpha$ is continuous at those points. Then, $f\in \mathcal{R}(\alpha )$ on $[a,b]$.

Proof
Note that again, it would suffice to consider the case where $f$ is discontinuous only at a single point in $[a,b]$, say $p$. Exactly the same set up as in the previous proof.

$$\begin{array}{rl}U(P,f,\alpha )-L(P,f,\alpha )=& \sum _{i=1}^{j-1}(M_i-m_i)\Delta {\alpha }_i\\ & +(M_j-m_j)\Delta {\alpha }_j\\ & +\sum _{i=j+1}^n(M_i-m_i)\Delta {\alpha }_i\end{array}$$

The fact that $M_j-m_j<2M$ remains. $\alpha$ being continuous implies $\Delta x_j\to 0$ $⟹$ $\Delta {\alpha }_j\to 0$. Thus, we can proceed to bound the expression on the right as we did in the previous proof. ❏

Rudin’s proof
Let ${ϵ}_1,{ϵ}_2>0$. Let $E$ be the set of points in $[a,b]$ where $f$ is discontinuous. Since $\alpha$ is continuous, we can construct intervals $[k_p,k_p^{\prime }]$ for each $p\in E$ such that ${\sum }_{p\in E}\alpha (k_p^{\prime })-\alpha (k_p)<{ϵ}_1$. Let $A={\bigcup }_{p\in E}(k_p,k_p^{\prime })$. The restriction of $f$ to $[a,b]\setminus A$ is uniformly continuous. Thus, we can choose a $\delta >0$ such that $s,t\in [a,b]\setminus A,|s-t|<\delta ⟹|f(s)-f(t)|<{ϵ}_2$. Now, construct a partition $P$ such that for all $p\in E$ there exists $i$ such that $x_{i-1}=k_p\in P$ and $x_i=k_p^{\prime }\in P$, and for all $x_i\ne k_p^{\prime }$ for any $p\in E$, $\Delta x_i<\delta$. As always, we have $M=\sup |f(x)|$.

$$\begin{array}{rl}U(P,f,\alpha )-L(P,f,\alpha )& <2M{ϵ}_1+{ϵ}_2(\alpha (b)-\alpha (a)-{ϵ}_1)\\ & <2M{ϵ}_1+{ϵ}_2(\alpha (b)-\alpha (a))\end{array}$$

Since ${ϵ}_1$ and ${ϵ}_2$ can be made arbitrarily small, it follows that $U(P,f,\alpha )-L(P,f,\alpha )<ϵ$ for all $ϵ>0$. ❏

Note that you could have made the same argument with a single $ϵ$. I choose to use two different variables to semantically indicate that they are bounding different quantities.

Compositions of integrable functions with continuous functions

Rudin, 6.11

Theorem 7(Theorem).

Suppose $f\in \mathcal{R}(\alpha )$ on $[a,b]$, $m\le f\le M$, $\varphi$ is continuous on $[m,M]$, and $h(x)=\varphi (f(x))$ on $[a,b]$. Then, $h\in \mathcal{R}(\alpha )$ on $[a,b]$.

Proof
Let $ϵ>0$. We have to show that there exists a partition $P^{\prime }$ of $[a,b]$ such that

$$\begin{array}{r}U(P^{\prime },h,\alpha )-L(P^{\prime },h,\alpha )<ϵ\end{array}$$

Since $\varphi$ is uniformly continuous on $[m,M]$, there exists $\delta >0$ such that $|s-t|<\delta$ $⟹$ $|\varphi (s)-\varphi (t)|<\boxed{{ϵ}_1}$ for any ${ϵ}_1>0$.

Since $f\in \mathcal{R}$ on $[a,b]$, there exists a partition $P=\{x_0,x_1,x_2,\dots ,x_n\}$ such that

$$\begin{array}{r}U(P,f,\alpha )-L(P,f,\alpha )<\boxed{{ϵ}_2}\end{array}$$

for any ${ϵ}_2>0$ (we will choose ${ϵ}_1$ and ${ϵ}_2$ to fit our needs later). Now,

$$\begin{array}{r}U(P,f,\alpha )-L(P,f,\alpha )=\sum _{i=1}^n(M_i-m_i)\Delta {\alpha }_i\end{array}$$

where $M_i=\sup f([x_{i-1},x_i])$ and $m_i=\inf f([x_{i-1},x_i])$. Let us try to use this partition $P$ for $h$ and see what comes out of it.

$$\begin{array}{r}U(P,h,\alpha )-L(P,h,\alpha )=\sum _{i=1}^n(M_i^{\prime }-m_i^{\prime })\Delta {\alpha }_i\end{array}$$

where $M_i^{\prime }=\sup h([x_{i-1},x_i])$ and $m_i^{\prime }=\inf h([x_{i-1},x_i])$. Notice that if $M_i-m_i<\delta$ for some $i$, $|f(x)-f(y)|<\delta$ for all $x,y\in [x_{i-1},x_i]$, so $|\varphi (f(x))-\varphi (f(y))|<{ϵ}_1$, i.e, $M_i^{\prime }-m_i^{\prime }<{ϵ}_1$. Of course, this is not going to happen for all $i$. Let the set $A$ contain all $i^{\prime }s$ for which $M_i-m_i<\delta$, and $B$ contain all $i$‘s for which $M_i-m_i\ge \delta$. Then,

$$\begin{array}{rl}\sum _{i=1}^n(M_i^{\prime }-m_i^{\prime })\Delta {\alpha }_i& =\sum _{i\in A}(M_i^{\prime }-m_i^{\prime })\Delta {\alpha }_i+\sum _{i\in B}(M_i^{\prime }-m_i^{\prime })\Delta {\alpha }_i\\ & <{ϵ}_1(\alpha (b)-\alpha (a))+\sum _{i\in B}(M_i^{\prime }-m_i^{\prime })\Delta {\alpha }_i\end{array}$$

Halfway there! Since we cannot make $M_i-m_i$ arbitrarily small for $i\in B$, let us instead focus on ${\sum }_{i\in B}\Delta {\alpha }_i$. Since $\delta \le M_i-m_i$, we have

$$\delta \sum _{i\in B}\Delta {\alpha }_i\le \sum _{i\in B}(M_i-m_i)\Delta {\alpha }_i<{ϵ}_2$$ $$⟹\sum _{i\in B}\Delta {\alpha }_i<\frac{{ϵ}_2}{\delta }$$

$\varphi$, being continuous on a closed interval, is bounded. Thus, we can put $M_i^{\prime }-m_i^{\prime }<K$ for all $i$ for some positive constant $K$. This gets us

$$\begin{array}{r}\sum _{i\in B}(M_i^{\prime }-m_i^{\prime })\Delta {\alpha }_i<\frac{K{ϵ}_2}{\delta }\end{array}$$

Thus, we have

$$\begin{array}{r}U(P,h,\alpha )-L(P,h,\alpha )<{ϵ}_1(\alpha (b)-\alpha (a))+\frac{K{ϵ}_2}{\delta }\end{array}$$

In hindsight, choosing ${ϵ}_2={\delta }^2$ would have been advantageous. Say we had done that. This gets us

$$U(P,h,\alpha )-L(P,h,\alpha )<{ϵ}_1(\alpha (b)-\alpha (a))+K\delta$$

In hindsight, choosing $\delta$ such that $\delta <{ϵ}_1$ would have been great. Let’s pretend we did that.

$$\begin{array}{r}U(P,h,\alpha )-L(P,h,\alpha )<{ϵ}_1(\alpha (b)-\alpha (a)+K)\end{array}$$

In hindsight, choosing ${ϵ}_1=ϵ/(\alpha (a)-\alpha (b)+K)$ would have been super smart. Say we were super smart. This finally gets us

$$U(P,h,\alpha )-L(P,h,\alpha )<ϵ$$

🎉🎉🎉 ❏

Info

The most general characterization of integrability goes like $f\in \mathcal{R}$ iff it is almost everywhere continuous.

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Properties of the integral

Rudin, 6.12

a

Theorem 8(Theorem).

If $f\in \mathcal{R}(\alpha )$, for every constant $c$ we have $cf\in \mathcal{R}(\alpha )$ and

$${\int }_a^{b}cfd\alpha =c{\int }_a^{b}fd\alpha .$$

Theorem 9(Theorem).

If $f_1\in \mathcal{R}(\alpha )$ and $f_2\in \mathcal{R}(\alpha )$ on $[a,b]$, then

$$f_1+f_2\in \mathcal{R}(\alpha ),$$ $${\int }_a^b{f}_1+f_{2}d\alpha ={\int }_a^b{f}_{1}d\alpha +{\int }_a^b{f}_{2}d\alpha .$$

Proof
Let $ϵ>0$. Find partitions $P_1$ and $P_2$ which satisfy the criterion for integrability for $f_1$ and $f_2$ respectively for $ϵ$. Let $P=P_1\cup P_2$ be their common refinement.

Note

For $f_1$ and $f_2$ defined on $X\subset \mathbb{R}$, we have

$$\underset{x\in X}{\inf }{f}_1(x)+\underset{x\in X}{\inf }{f}_1(x)\le \underset{x\in X}{\inf }(f_1+f_2)(x)$$

Proof:
For any $x\in X$ we have

$$\underset{t\in X}{\inf }{f}_1(t)\le f_1(x)\underset{t\in X}{\inf }{f}_2(t)\le f_2(x)$$

Adding these we get

$$\underset{t\in X}{\inf }{f}_1(t)+\underset{t\in X}{\inf }{f}_2(t)\le (f_1+f_2)(x)$$

Taking the infimum over all $x$ on the RHS, we get our desired result.

Using the above result we can write

$$\begin{array}{r}U(P,f_1+f_2,\alpha )\le U(P,f_1,\alpha )+U(P,f_2,\alpha ),\\ L(P,f_1+f_2,\alpha )\ge L(P,f_1,\alpha )+L(P,f_2,\alpha ).\end{array}$$

Thus, we have

$$\begin{array}{rl}U(P,f_1+f_2,\alpha )-L(P,f_1+f_2,\alpha )\le & U(P,f_1,\alpha )-L(P,f_1,\alpha )+\\ & U(P,f_2,\alpha )-L(P,f_2,\alpha )\\ <& 2ϵ.\end{array}$$

Therefore, $f_1+f_2\in \mathcal{R}(\alpha )$.

For the same $P$, we have

$$U(P,f_1,\alpha )<{\int }_a^b{f}_{1}d\alpha +ϵ.$$

A similar equation exists for $f_2$. Thus, we have

$${\int }_a^b{f}_1+f_{2}d\alpha \le U(P,f_1+f_2,\alpha )<{\int }_a^b{f}_{1}d\alpha +{\int }_a^b{f}_{2}d\alpha +2ϵ.$$

❏

b

Theorem 10(Theorem).

If $f_1\in \mathcal{R}(\alpha )$ and $f_2\in \mathcal{R}(\alpha )$ on $[a,b]$ and $f_1(x)\le f_2(x)$, then

$${\int }_a^b{f}_{1}d\alpha \le {\int }_a^b{f}_{2}d\alpha .$$

c

Theorem 11(Theorem).

If $f\in \mathcal{R}(\alpha )$ on $[a,b]$ and if $a<c<b$, then $f\in \mathcal{R}(\alpha )$ on $[a,c]$ and on $[c,b]$, and

$${\int }_a^{c}fd\alpha +{\int }_c^{b}fd\alpha ={\int }_a^{b}fd\alpha .$$

Proof
Let $ϵ>0$. Pick partition $P$ of $[a,b]$ such that $c\in P$ and $U(P,f,\alpha {)}_{[a,b]}-L(P,f,\alpha {)}_{[a,b]}<ϵ$. The definitions of $U$ and $L$ tell us that this implies

$$\begin{array}{rl}& (U(P,f,\alpha {)}_{[a,c]}-L(P,f,\alpha {)}_{[a,c]})+\\ & (U(P,f,\alpha {)}_{[c,b]}-L(P,f,\alpha {)}_{[c,b]})<ϵ\end{array}$$

where the elements of $P$ not in the pertinent interval are ignored. Since both quantities are positive, each one must be less than $ϵ$. Thus, $f\in \mathcal{R}(\alpha )$ on $[a,c]$ and $[c,b]$. Again,

$${\int }_a^{b}fd\alpha \le U(P,f,\alpha {)}_{[a,c]}+U(P,f,\alpha {)}_{[c,b]}<{\int }_a^{c}fd\alpha +{\int }_c^{b}fd\alpha +2ϵ$$

❏

d

Theorem 12(Theorem).

If $f\in \mathcal{R}(\alpha )$ on $[a,b]$ and if $|f(x)|\le M$ on $[a,b]$, then

$$\left|{\int }_a^{b}fd\alpha \right|\le M[\alpha (b)-\alpha (a)].$$

e

Theorem 13(Theorem).

If $f\in \mathcal{R}(\alpha )$ and $c$ is a positive constant, then $f\in \mathcal{R}(c\alpha )$ and

$${\int }_a^{b}fd(c\alpha )=c{\int }_a^{b}fd\alpha .$$

Theorem 14(Theorem).

If $f\in \mathcal{R}({\alpha }_1)$ and $f\in \mathcal{R}({\alpha }_2)$, then $f\in \mathcal{R}({\alpha }_1+{\alpha }_2)$, and

$${\int }_a^{b}fd({\alpha }_1+{\alpha }_2)={\int }_a^{b}fd{\alpha }_1+{\int }_a^{b}fd{\alpha }_2.$$