# Math4121 Lecture 39

## Fundamental theorem of calculus (In Lebesgue integration)

### Preliminary results

#### Lemma 1

Riemann integrable functions are Lebesgue integrable
$$\int_a^b f(x)dx = \int_{[a,b]} f dm$$

#### Lemma 2

Density of continuous functions: Given $f$ integrable, then $\exists \epsilon > 0$ there is $g$ continuous such that $\int_{[a,b]} |f-g| dm < \epsilon$

#### Lemma 3

Maximal function: $f^*(x) = \sup_{I\text{ is open intervals}}A_I f(x)$, where $A_I = \frac{\chi_I}{m(I)} \int_I f dm$. Then $|\{x\in\mathbb{R}:f^*(x)>\lambda\}|<\frac{2}{\lambda}\int_{\mathbb{R}}|f|dm$

#### Lemma 4

$I=[a,b]$, $I_\delta = [a+\delta, b-\delta]$, $\delta>0$, $\lim_{\delta\to 0^+} A_{I_\delta} f(x) = A_I f(x)$. (Prove via dominated convergence theorem)

> Riemann's Fundamental theorem of calculus: 
> 
> If $g$ is continuous on $[a,b]$, then $G(x) = \int_a^x g(t)dt$ is differentiable on $(a,b)$ and $G'(x) = g(x)$ for all $x\in(a,b)$.

### Lebesgue's Fundamental theorem of calculus

If $f$ is Lebesgue integrable on $[a,b]$, then $F(x) = \int_a^x f(t)dt$ is differentiable almost everywhere and $F'(x) = f(x)$ almost everywhere.

Outline:

Let $\lambda,\epsilon > 0$. Find $g$ continuous such that $\int_{\mathbb{R}}|f-g|dm < \frac{\lambda \epsilon}{5}$.

To control $A_I f(x)-f(x)=(A_I(f-g)(x))+(A_I g(x)-g(x))+(g(x)-f(x))$, we need to estimate the three terms separately.

Our goal is to show that $\lim_{r\to 0^+}\sup_{I\text{ is open interval}, m(I)<r, x\in I}|A_I f(x)-f(x)|=0$. For $x$ almost every $x\in[a,b]$.

This implies the fundamental theorem of calculus.

Since $\frac{F(x+h)-F(x)}{h}=\frac{1}{m(I_h)}\int_{I_h}f dm$, if the above condition holds, then $\forall \eta>0$, we can find $r>0$ such that $\sup_{I\text{ is open interval}, m(I)<r, x\in I}|A_I f(x)-f(x)|<\frac{\eta}{2}$.

Now given $h<\min\{r, x-a\}$, we can find by [4](https://notenextra.trance-0.com/Math4121_L39.html#Lemma-4) an interval $I_h^*$ such that

$$
\left|\frac{1}{h}\int_{I_h^*}f dm - \frac{1}{m(I_h^*)}\int_{I_h^*}f dm\right|<\frac{\eta}{2}
$$

Proof:

Let

$$
F=\left\{x\in [a,b]:\limsup_{r\to 0^+}\sup_{I\text{ is open interval}, m(I)<r, x\in I}|A_I f(x)-f(x)|>\lambda \right\}
$$

Need to show $m(F)<\epsilon$.

Since $F\subseteq \{(f-g)^*>\frac{\lambda}{2}\}\cup \{(f-g)>\frac{\lambda}{2}\}$

$$
\limsup_{r\to 0^+}\sup_{I\text{ is open interval}, m(I)<r, x\in I}|A_I f(x)-f(x)|\leq \limsup_{r\to 0^+}\sup_{I\text{ is open interval}, m(I)<r, x\in I}|A_I (f-g)(x)|+|g(x)-f(x)|
\leq |(f-g)^*(x)|+|(f-g)(x)|
$$

By maximal inequality and Markov's inequality,

$$
\begin{aligned}
m(F)&\leq \frac{4}{\lambda}\int_{\mathbb{R}}|f-g|dm+\frac{1}{\lambda}\int_{\mathbb{R}}|f-g|dm\\
&=\frac{5}{\lambda}\frac{\lambda \epsilon}{5}\\
&=\epsilon
\end{aligned}
$$

QED
