Proof of [tex]\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A[/tex]

[3.1415926535]

\text{ Let }f:\mathbb{R}\to [0,\infty] \text{ be a measurable function and }A\subset \mathbb{R}Then, show that
\begin{equation}
\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A \tag{1}
\end{equation}
\text{ where ${1}_A$ is the characteristic function of $A$ defined as }
\begin{equation}
{1}_A(x)=\begin{cases}1 & \text{if $x\in A$,}
\\
0 &\text{if $x\notin A$.}
\end{cases} \tag{2}
\end{equation}
\text{ and $\int\limits_{A}f$ is the Lebesgue integral of $f$ on $A$ defined as:}
\begin{equation}
\int\limits_{A}f=\sup\left\{\int\limits_{A}s:0\le s\le f\text{ and }s\text{ is simple}\right\} \tag{3}
\end{equation}

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

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

\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)?

 

[3.1415926535]

PROOF: http://math.stackexchange.com/quest...-int-limits-mathbbrf1-a-for-the-lebesgue-inte

 

[Undecided Guy]

The statement can be easily shown via the MCT.

 

[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.

 

[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.