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I am trying to show the following:

Let $f$ be a measurable function on a set $E$, and let $A$ be a measurable subset of $E$. Then, using the definition of the Lebesgue Integral of nonnegative functions (given below), prove that $\int_{A}f=\int_{E}\chi_{A}f$.

Definition of the Lebesgue Integral for Nonnegative Functions: For an arbitrary nonnegative measurable function $f:E\to \mathbb{R} \cup + \infty$, define its Lebesgue integral by $\int_{E}f = \sup \left\{\int_{E} h \vert h\,\text{bounded, measurable, of finite support, and}\,0 \leq h \leq f\, \text{on}\,E \right\}$

Now, the definition holds only for nonnegative measurable functions. So, I let $f = f^{+}-f^{-}$ on $E$. Then, we consider $\int_{E}\chi_{A}f = \int_{E}\chi_{A}f^{+}-\int_{E}\chi_{A}f^{-} $.

Applying the definition, we get that this is equal to $\sup\{\int_{E}h\vert h\,\text{bounded, measurable, of finite support, and}\,0 \leq h \leq \chi_{A}f^{+}\,\text{on}\,E\} -\sup\{\int_{E}h\vert \tilde{h}\,\text{bounded, measurable, of finite support, and}\,0 \leq \tilde{h} \leq \chi_{A}f^{+}\,\text{on}\,E\}$

where $h$ and $\tilde{h}$ are suitable nonnegative, bounded, measurable functions with finite support.

Since $A \subseteq E$, $h \leq \chi_{A}f^{+}$ and $\tilde{h} \leq \chi_{A}f^{-}$ on $A$ as well, and since $\chi_{A}$ is the characteristic function of the set $A$, $\chi_{A}f^{+}=\chi_{A}f^{-}=0$ on $E\backslash A$, so $h = \tilde{h}=0$ there as well.

Now, there is a corollary to the Monotone Convergence Theorem I would like to use in order to help me introduce series in a similar manner as in the second answer to this question (proving the same thing, but instead using the definition of the Lebesgue Integral for bounded functions). The corollary states as follows:

Let $\{ u_{n}\}$ be a sequence of nonnegative measurable functions on $E$. If $f=\sum_{n=1}^{\infty}u_{n}$ pointwise a.e. on $E$, then $\int_{E}f=\sum_{n=1}^{\infty}\int_{E}u_{n}$.

I'd somehow like to be able to express both $h$ and $\tilde{h}$ as a sum $\displaystyle \sum_{i=1}^{n} u_{i}\chi_{\displaystyle [E_{i}\backslash A) \cup (E_{i}\cap A)]}$. But I am not sure I am 1) justified in doing it, 2) exactly how to do it.

Then, once I have that, I can apply the Corollary to turn the finite sum into an infinite series, and can use a bunch of bounding to get the result that I want. I just require assistance with this one part.

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That seems an unnecessary approach here. First prove the theorem for non-negative functions, then extend to all functions as you've indicated.

For nonnegative $f$, note that if $h$ is bounded and has finite support on $A$, then you can define $$h_E(x) =\begin{cases} h(x) & x\in A\\0 & x \in E \setminus A\end{cases}$$ Then $h_E$ is bounded and has finite support on $E$. Further, $h_E\chi_A = h_E$ and $h_E|_A = h$. Conversely if $h$ is bounded and has finite support on $E$, then $h\chi_A|_A$ is bounded and has finite support on $A$.

Edit: as requested, where I've tried to dot all the i's and cross all the t's.

For notational convenience, for measurable sets $B$ and measurable non-negative functions $g$, define ${\scr M}(g, B)$ to be the set of all bounded measurable functions $h$ of finite support on $B$ such that $0 \le h \le g$.

Let $f$ be a non-negative measurable function on $E$ and $A$ a measurable subset of $E$. Then $\chi_Af$ is also a measurable function on $E$.

Now if $h \in {\scr M}(\chi_Af,E)$ and $x \in E \setminus A$, then $0 \le h(x) \le \chi_A(x)f(x) = 0\cdot f(x) = 0$. Hence $\operatorname{supt} h \subseteq A$. And for $x \in A$, $0 \le h(x) \le \chi_A(x)f(x) = 1\cdot f(x) = f(x)$, so $0 \le h \le f$ on $A$. Therefore $h|_A \in {\scr M}(f,A)$. And, $$\int_E h = \int_A h + \int_{E\setminus A} h = \int_A h$$ since $h = 0$ on $E\setminus A$.

Conversely, if $h' \in {\scr M}(f,A)$, then define $h : E \to \bar{\Bbb R}$ by $$h(x) = \begin{cases} h'(x) & x \in A\\0 & x \notin A\end{cases}$$ Then $h$ is measurable, $\operatorname{supt} h = \operatorname{supt} h'$ and therefore is finite, and if $h' < M$, then $M > 0$, so $h < M$ as well. And lastly $$h(x) = \begin{cases} h'(x) \le f(x) = \chi_A(x)f(x) & x \in A\\0 = \chi_A(x)f(x) & x \notin A\end{cases}$$ Hence $h \in {\scr M}(\chi_Af,E), h' = h|_A$ and therefore $$\int_E h = \int_A h'.$$ Therefore the restriction map $T\ :\ {\scr M}(\chi_Af,E) \to {\scr M}(f,A)\ :\ h \to h_A$ is a bijection, and $\int_E h = \int_A T(h)$. So $$\begin{align}\int \chi_Af &= \sup\left\{\int_E h \ :\ h \in {\scr M}(\chi_Af,E)\right\}\\ &=\sup\left\{\int_A T(h) \ :\ h \in {\scr M}(\chi_Af,E)\right\}\\ &=\sup\left\{\int_A T(h) \ :\ T(h) \in {\scr M}(f,A)\right\}\\ &=\sup\left\{\int_A h' \ :\ h' \in {\scr M}(f,A)\right\}\\ &=\int_A f\end{align}$$

When $f$ may be negative, $$\int_E\chi_Af = \int_E\chi_Af^+ -\int_E\chi_Af^- = \int_A f^+ - \int_Af^- = \int_Af$$

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  • $\begingroup$ if you add more details to this answer, I will award you the bounty. It's an excellent solution, but I want to see more. $\endgroup$ – ALannister Nov 18 '15 at 15:10
  • $\begingroup$ @JessyCat - see the edit. $\endgroup$ – Paul Sinclair Nov 19 '15 at 0:52

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