[tex]\text{ Let }f:\mathbb{R}\to [0,\infty] \text{ be a measurable function and }A\subset \mathbb{R}[/tex]Then, show that(adsbygoogle = window.adsbygoogle || []).push({});

\begin{equation}

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

\end{equation}

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

\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}

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

\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}

[tex]\text{where $s:\mathbb{R}\to [0,\infty]$ is a simple function. Thus to prove (1) we need to show that:}[/tex]

\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)?

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

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