# Proof of $$\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$$

1. Jun 1, 2012

### 3.1415926535

$$\text{ Let }f:\mathbb{R}\to [0,\infty] \text{ be a measurable function and }A\subset \mathbb{R}$$Then, show that

\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A \tag{1}

$$\text{ where {1}_A is the characteristic function of A defined as }$$

{1}_A(x)=\begin{cases}1 & \text{if $x\in A$,}
\\
0 &\text{if $x\notin A$.}
\end{cases} \tag{2}

$$\text{ and \int\limits_{A}f is the Lebesgue integral of f on A defined as:}$$

\int\limits_{A}f=\sup\left\{\int\limits_{A}s:0\le s\le f\text{ and }s\text{ is simple}\right\} \tag{3}

I can easily prove this property for simple functions so take this for granted:

\int\limits_{A}s=\int\limits_{\mathbb{R}}s{1}_A \tag{4}

$$\text{where s:\mathbb{R}\to [0,\infty] is a simple function. Thus to prove (1) we need to show that:}$$
\begin{gather}
\sup\left\{\int\limits_{A}s:0\le s\le f\text{ and }s\text{ is simple}\right\}=\sup\left\{\int\limits_{\mathbb{R}}s:0\le s\le f{1}_A\text{ and }s\text{ is simple}\right\}\notag\\
\sup\left\{\int\limits_{\mathbb{R}}s{1}_A:0\le s\le f\text{ and }s\text{ is simple}\right\}=\sup\left\{\int\limits_{\mathbb{R}}s:0\le s\le f{1}_A\text{ and }s\text{ is simple}\right\} \tag{5}
\end{gather}

My question is how do we prove (5)?

2. Jun 1, 2012

### 3.1415926535

Last edited: Jun 1, 2012
3. Jun 2, 2012

### Undecided Guy

The statement can be easily shown via the MCT.

4. Jun 4, 2012

### Undecided Guy

Specifically, I was thinking something like this:

the statement is obvious for simple functions. Let f be some nonnegative measurable function on $\mathbb{R}$. Construct a sequence of simple functions $\{s_n\}_{n=1}^{\infty}$ so that $s_n(x) \uparrow f(x)$ as $n \to \infty$. Then we also have that $(s_n 1_A)(x) \uparrow (f 1_A)(x)$.

By the monotone convergence theorem, we have that $\int_A s_n \to \int_A f$ as $n \to \infty$ and that $\int_{\mathbb{R}} s_n 1_A \to \int_{\mathbb{R}} f 1_A$ as $n \to \infty$.

But we know that for each n, $\int_{\mathbb{R}} s_n 1_A = \int_A s_n$. So we have that $\int_A s_n \to \int_{\mathbb{R}} f 1_A$. But limits of real sequences are unique, so it follows that $\int_A f = \int_{\mathbb{R}} f 1_A$.

5. Jun 5, 2012

### deluks917

Let L(f) = ∫Af-∫ℝf1A. We note L(s) = 0 for all step maps. Since L is continuous in the L1 norm and step maps are dense in L1 we have that L(f) = 0 for all f.