3.31 Integrable Limit theorem

We begin the lecture of today with a theorem that states that given a continuous function $f: [a, b] \rightarrow \mathbb{R}$, its integral can be computed as a mean value, that is there exists a point $c \in [a, b]$ such that the integral is equal to $f(c)(b - a)$.

Since $f$ is a continuous function on a closed interval, it is bounded and there exists $m, M \in \mathbb{R}$ such that $m \leq f(x) \leq M \quad \forall x \in [a, b]$

By the theorem we saw in the last lecture, this implies that

\[ \underbrace{m(b-a)}_{\text{area of blue rectangle}}\leq\underbrace{\int_{a}^{b}f(x)dx} _{\exists c\text{ such that }=f(c)(b-a)}\leq \underbrace{M(b-a)}_{\text{area of big red rectangle}} \]

Theorem

Let $f: [a, b] \rightarrow \mathbb{R}$ be a continuous function. Then there exists a point $c \in [a, b]$ such that

\[ \int_{a}^{b}f(x)dx=f(c)(b-a) \]

Proof

Since $f$ is continuous in $[a,b]$, it has a minimum and maximum, that is $\exists y,z\in[a,b]:f(y)=m\leq f(x)\leq M=f(z)$

By bounded function property, $\displaystyle m(b - a) \leq \int_{a}^{b}f(x)dx \leq M(b - a)$

Then $\displaystyle f(y)\leq \frac{1}{b-a}\int^{b}_{a}f(x)dx\leq f(z)$. Note that $\displaystyle{\frac{1}{b-a}\int_{a}^{b}f(x)dx}$ is a number $\in \R$

Then recall Intermediate Value Theorem, $\exists c\in [a,b]:f(c)=\displaystyle{\frac{1}{b-a}\int_{a}^{b}f(x)dx}$

Thus ${\displaystyle{\int_{a}^{b}f(x)dx}}=f(c)(b-a)$

Example

$f:[0,2]\to \R,f(x)=x$ is continuous and $a=0,b=2$, then ${\displaystyle \int^{2}_{0}xdx}=\frac{2^{2}}{2}=2$

Then $2=f(c)(b-a)=f(c)\cdot 2$, thus $f(c)=1$, then $c=1$

Next we analyze the relation between limits of sequences of functions and integration.

$f_n(x) = \begin{cases} n & \text{if } 0 < x < \frac{1}{n} \\ 0 & \text{if } x = 0 \text{ or } \frac{1}{n} \le x \le b \end{cases}$ and $f_{n}\xrightarrow{\text{pointwise}}f=0$ in $[a, b]$, then $\displaystyle \int_{a}^{b} f_{n}(x) dx \xrightarrow{?}\int_{a}^{b} f(x) dx$

$\displaystyle\int_{0}^{1} f_{3}(x) dx = 1$, $\displaystyle\int_{0}^{1} f_{2}(x) dx = 1$, $\displaystyle\int_{0}^{b} f_{1}(x) dx = 1$

But we know $f_n(x) \to 0$ and $\displaystyle\int_{a}^{b} f_{n}(x) dx \to 1$

Thus pointwise cannot imply this

Theorem (Integrable Limit theorem)

Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of integrable functions on an interval $[a, b]$ with $b > a$.

If $(f_n)_{n \in \mathbb{N}}$ converges uniformly to a function $f$ on $[a, b]$, then

\[ f\text{ is integrable on }[a, b]\text{ and }\displaystyle\int_{a}^{b}f_{n}\rightarrow \int_{a}^{b}f \]

Proof

Let's $U(f)=\inf\limits_{P\in\mathcal{P}}\left\lbrace U\left(f,P\right)\right\rbrace,U( f_{n})=\inf\limits_{P\in\mathcal{P}}\left\lbrace U\left(f_{n},P\right)\right\rbrace ,L(f)=\sup\limits_{P\in\mathcal{P}}\left\lbrace L\left(f,P\right)\right\rbrace,L( f_{n})=\sup\limits_{P\in\mathcal{P}}\left\lbrace L\left(f_{n},P\right)\right\rbrace$

We want to show $U(f)=L(f)$ and we know $U(f_n)=L(f_n)$ since $f_n$ is integrable

Also, $f_n\xrightarrow {u}f$, so given $\varepsilon>0$, there exists $N\in\mathbb{N}:|f_{n}(x)-f(x)|<\frac{\varepsilon}{2\left(b-a\right)},\forall x\in [a,b]$

Hence $-\frac{\varepsilon}{2\left(b-a\right)}<f\left(x\right)-f_{n}\left(x\right)<\frac{\varepsilon}{2\left(b-a\right)},\forall x\in[a,b] $

Then $f_{n}(x)-\frac{\varepsilon}{2\left(b-a\right)}<f\left(x\right)<f_{n}\left(x\right )+\frac{\varepsilon}{2\left(b-a\right)},\forall x\in[a,b]$

Given a partition $P=(x_{0}=a,x_{1},...,x_{n}=b)$, we have that $f_{n}(x)-\frac{\varepsilon}{2\left(b-a\right)}<f\left(x\right)<f_{n}\left(x\right )+\frac{\varepsilon}{2\left(b-a\right)},\forall x\in[x_{i-1},x_{i}]$

Then $\inf\limits_{x\in[x_{i-1},x_{i}]}\left\lbrace f_{n}\left(x\right)-\frac{\varepsilon}{2\left(b-a\right)} \right\rbrace\leq\inf\limits_{x\in[x_{i-1},x_{i}]}\left\lbrace f\left(x\right)\right \rbrace\leq\inf\limits_{x\in[x_{i-1},x_{i}]}\left\lbrace f_{n}\left(x\right)+\frac{\varepsilon}{2\left(b-a\right)} \right\rbrace$

Then $\inf\left\lbrace f_{n}\left(x\right):x\in[x_{i-1},x_{i}]\right\rbrace-\frac{\varepsilon}{2\left(b-a\right)} \leq\inf\left\lbrace f\left(x\right):x\in[x_{i-1},x_{i}]\right\rbrace\leq\inf\left \lbrace f_{n}\left(x\right):x\in[x_{i-1},x_{i}]\right\rbrace+\frac{\varepsilon}{2\left(b-a\right)}$

Then $m_{i}^{(n)}-\frac{\varepsilon}{2\left(b-a\right)}\leq m_{i}\leq m_{i}^{\left(n\right)} +\frac{\varepsilon}{2\left(b-a\right)},\forall i$

Then $\sum_{i=1}^{n}\left(m_{i}^{\left(n\right)}-\frac{\varepsilon}{2\left(b-a\right)} \right)\left(x_{i}-x_{i-1}\right)\leq\sum_{i=1}^{n}m_{i}\left(x_{i}-x_{i-1}\right )\leq\sum_{i=1}^{n}\left(m_{i}^{\left(n\right)}+\frac{\varepsilon}{2\left(b-a\right)} \right)\left(x_{i}-x_{i-1}\right)$

Then $\sum_{i=1}^{n}m_{i}^{\left(n\right)}\left(x_{i}-x_{i-1}\right)-\frac{\varepsilon}{2\left(b-a\right)} \sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)\leq L\left(f,P\right)\leq\sum_{i=1}^{n}m _{i}^{\left(n\right)}\left(x_{i}-x_{i-1}\right)+\frac{\varepsilon}{2\left(b-a\right)} \sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)$

Then $L(f_{n},P)-\frac{\varepsilon}{2}\leq L\left(f,P\right)\leq L\left(f_{n},P\right) +\frac{\varepsilon}{2}$

Then $\sup\limits_{P\in\mathcal{P}}\left\lbrace L(f_{n},P)-\frac{\varepsilon}{2}\right \rbrace\leq\sup\limits_{P\in\mathcal{P}}\left\lbrace L\left(f,P\right)\right\rbrace \leq\sup\limits_{P\in\mathcal{P}}\left\lbrace L\left(f_{n},P\right)+\frac{\varepsilon}{2} \right\rbrace$

Then $\sup\limits_{P\in\mathcal{P}}\left\lbrace L(f_{n},P)\right\rbrace-\frac{\varepsilon}{2} \leq\sup\limits_{P\in\mathcal{P}}\left\lbrace L\left(f,P\right)\right\rbrace\leq\sup \limits_{P\in\mathcal{P}}\left\lbrace L\left(f_{n},P\right)\right\rbrace+\frac{\varepsilon}{2}$

Then $L(f_{n})-\frac{\varepsilon}{2}\leq L\left(f\right)\leq L\left(f_{n}\right)+\frac{\varepsilon}{2}$, thus $|L(f)-L(f_{n})|\leq\frac{\varepsilon}{2}$

Similarly, $|U(f)-U(f_{n})|\leq\frac{\varepsilon}{2}$ (*)

Thus $|U(f)-L(f)|\leq|U(f)-U(f_{n})|+|U(f_{n})-L(f)|=|U(f)-U(f_{n})|+|L(f_{n})-L(f)|\leq \varepsilon$

Thus $U(f)=L(f)$ and $f$ is integrable

Thus $\displaystyle \int^b_af=U(f)=L(f)$, also from (*) we have $\lim_{n\to\infty}U(f_n)=U(f)$

Thus $U(f_n)=\displaystyle \int^b_af_n\to \int ^b_af$

Remark

$\sum\limits_{n=0}^{\infty}a_{n}x^{n}= f(x)$ on $(-R, R)$

$\Rightarrow\sum\limits_{n=0}^{N}a_{n}x^{n}\xrightarrow{u}f(x)$ on $[a,b] \subseteq (-R, R)$

$\displaystyle\Rightarrow f(x) = \sum_{n=0}^{\infty}a_{n} x^{n}$ is integrable on $[a,b]$ and $\displaystyle\lim_{N \to \infty}\sum_{n=0}^{N}a_{n}\int_{a}^{b}x^{n}dx = \int_{a} ^{b}f(x) dx$

Thus we have ${\displaystyle\sum_{n=0}^{\infty}a_{n}\int_{a}^{b}x^{n}dx=\int_{a}^{b}\sum\limits_{n=0}^{\infty}a_{n}x^{n}dx}$

Example

$f(x) = \frac{1}{1-x}= \sum\limits_{n=0}^{\infty}x^{n}$ in $(-1, 1)$

If $[a, b] \subset (-1, 1)$, we compute ${\displaystyle\int_{a}^{b}}\frac{1}{1-x}dx$ using the limit of $\displaystyle\int_{a}^{b} (\sum_{n=0}^{\infty}x^{n}) dx = \sum_{n=0}^{\infty}\int_{a}^{b} x^{n} dx$, How do we compute this?

For $[0,t]$, $0 < t < 1$

$\displaystyle\int_{0}^{t}\frac{1}{1-x}dx = \sum_{n=0}^{\infty}\int_{0}^{t}x^{n}d x = \sum_{n=0}^{\infty}\frac{t^{n+1}}{n+1}= \sum_{n=1}^{\infty}\frac{t^{n}}{n}= - \ln(1-t)$
$f(x) = e^x = \exp(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ in $\mathbb{R}$

In $[a, b] \subset \mathbb{R}$ we have ${\displaystyle\int_{a}^{b}\left(\sum_{n=0}^{N}\frac{x^{n}}{n!}\right)dx\rightarrow\int_{a}^{b}e^{x}dx}$

For $[0, t] \subset \mathbb{R}$, $\displaystyle\int_{0}^{t}(\sum_{n=0}^{N}\frac{x^{n}}{n!}) dx = \sum_{n=0}^{N}(\frac{1}{n!} \int_{0}^{t}x^{n}dx) = \sum_{n=0}^{N}\frac{1}{n!}\frac{t^{n+1}}{n+1}= \sum_{n=0}^{N} \frac{t^{n+1}}{(n+1)!}$

$\displaystyle= \sum_{n=0}^{N}\frac{t^{n}}{n!}- 1 \xrightarrow{N \to \infty}e^{t} - 1 \Rightarrow \int_{0}^{t}e^{x}dx = e^{t}- 1$

Definition

Let $f: [a, b] \rightarrow \mathbb{R}$ be an integrable function. For any $a \leq x \leq b$, we define the function $G: [a, b] \rightarrow \mathbb{R}$ by $\displaystyle G(x) = \int_{a}^{x}f(t)dt \quad \forall a \leq x \leq b$

In particular, we have that $\displaystyle G(a)=\int_{a}^{a}f(t)dt=0,G(b)=\int_{a}^{b}f(t)dt$

Theorem

Let $f: [a, b] \rightarrow \mathbb{R}$ be an integrable function and define $G: [a, b] \rightarrow \mathbb{R}$ as above. Then $G$ is continuous on $[a, b]$.

Proof

To prove $G$ is continuous, NTP: $\lim_{h\to 0} G(c+h)=G(c)$

Let's assume first that $h>0$, then $a<c<c+h<b$

$G(c+h)=\displaystyle\int_{a}^{c+h}f(t)dt=\int_{a}^{c}f(t)dt+\int_{c}^{c+h}f(t) dt$ and $G(c)=\displaystyle \int^c_af(t)dt$

So $|G(c+h)-G(c)|={\displaystyle\left|\int_{c}^{c+h}f\left(t\right)dt\right|}$

Since $f$ is integrable, $f$ is bounded(this is because the definition of integrable is started from closed interval), that is $\exists M>0,M\in \R$ such that $|f(x)|<M$

From $|G(c+h)-G(c)|<\displaystyle \int_{c}^{c+h}Mdt=M(c+h-c)=Mh$

When $h\to0$, $|G(c+h)-G(c)|\to 0$

Exercise, do the proof of $h<0$

Remark

As we have seen in the examples above, there is a closed relation between the integral of a function and the values of a primitive on the extremal points of the interval.

This is a general fact given by the following theorem, which will help us to compute integrals using primitives.