# Math4121 Lecture 37

## Extended fundamental theorem of calculus with Lebesgue integration

### Density of continuous functions

#### Lemma for density of continuous functions

Let $K\subseteq U$ be bounded sets in $\mathbb{R}$, $K$ is closed and $U$ is open. Then there is a continuous function $g$ such that $\chi_K\leq g\leq \chi_U$.

Proof in homework.

Hint: Consider the basic intervals cases.

#### Theorem for continuous functions

Let $f$ be integrable. For each $\epsilon>0$, there is a continuous function $g:\mathbb{R}\to\mathbb{R}$ such that $\int_{\mathbb{R}}|f-g|dm<\epsilon$.

Proof:

First where $f=\chi_S$ for some bounded means set $S$. then extended to all $f$ integrable.

First, assume $f=\chi_S$. Let $\epsilon>c$, we can find $K\subseteq S\subseteq U$. and $K$ is closed and $U$ is open such that (by definition of Lebesgue outer measure)

$$
m(K)+\frac{\epsilon}{2}>m(S)>m(U)-\frac{\epsilon}{2}
$$

In particular, $m(U\setminus K)=m(U)-m(K)<\epsilon$.

By lemma, there is a continuous function $g$ such that $\chi_K\leq g\leq \chi_U$.

So

$$
\int_E |\chi_S -g|dm=\int_{U\setminus K} |\chi_S -g|dm\leq m(U\setminus K)<\epsilon
$$

For the general case,

By the Monotone Convergence Theorem (use $|f|\chi_{[-N,N]}$ to approximate $|f|$), we can find $N$ large such that

$$
\int_{E_N^c}|f|dm<\frac{\epsilon}{3}
$$

where $E_N=E\cap [-N,N]$.

Notice that by the definition of Lebesgue integral, $\int f^+ dm=\sup\{\int \phi^+ dm:\phi\text{ is simple and } \phi\leq f^+\}$ and $\int f^- dm=\sup\{\int \phi^- dm:\phi\text{ is simple and } \phi\leq f^-\}$.

By considering $f^+$ and $f^-$ separately, we can find a simple function $\phi$ such that

$$
\int_{E_N} |f-\phi|dm<\frac{\epsilon}{3}
$$

For each $i=1,2,\cdots,n$, we can find $g_i$ continuous such that

$$
\int_{E}|\chi_{S_i}-g_i|dm<\frac{\epsilon}{3M}
$$

where $M=\sum_{i=1}^n |\alpha_i|$.

Take $g=\sum_{i=1}^n \alpha_i g_i$,

$$
\int_E |\phi-g|dm\leq \sum_{i=1}^n |\alpha_i|\int_E |g_i-\chi_{S_i}|dm<\frac{\epsilon}{3}
$$

$\phi-g=\sum_{i=1}^n \alpha_i (\chi_{S_i-g_i})$

All in all,

$$
\begin{aligned}
\int_E |f-g|dm&\leq \int_E|f-\phi|dm+\int_E |\phi-g|dm\\
&=\int_{E_N^c}|f|dm+\int_E |f-\phi|dm+\int_E |\phi-g|dm\\
&<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}\\
&=\epsilon
\end{aligned}
$$

QED

### Road map for proving the fundamental theorem of calculus in Lebesgue integration

Recall the Riemann-Stieltjes integral:

If $g\in \mathscr{R}(\alpha)$ on $[a,b]$,

$G(x)=\int_a^x g d\alpha$,

then:

1. $G$ is continuous on $[a,b]$
2. If $g$ is continuous at $x\in [a,b]$, then $G$ is differentiable at $x$ with $G'(x)=g(x)$.

To extend this to the case where $g$ is Lebesgue integrable, we use the Hardy-Littlewood maximal function.

#### Definition of the Hardy-Littlewood maximal function

Given an interval $I\subseteq \mathbb{R}$, define the averaging operator $A_I f(x)=\frac{\chi_I(x)}{m(I)}\int_I f(x)dm$.

(This function takes the average of $f$ over the interval $I$.)

The Hardy-Littlewood maximal function is defined as:

$$
f^*(x)=\sup_{I\text{ is open interval}}A_I f(x)
$$

We will show that $f^*$ is not that such worse than $f$. (Prove on Wednesday)

Relates to the Fundamental Theorem of Calculus in Lebesgue integration.

$$
\frac{G(x+h)-G(x)}{h}=\frac{1}{h}\int_x^{x+h} g(t)dt=A_{[x,x+h]}g(x)
$$

If we can control all the averages, we can control the function.