# Math4121 Lecture 37

## Extended fundamental theorem of calculus with Lebesgue integration

### Density of continuous functions

#### Lemma: 

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

$$
\int_{E_N^c}|f|dm<\frac{\epsilon}{2}
$$

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

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




QED