# Proving some points missed in my proof.

The question was:(From Royden "Real Analysis" fourth edition)

Let $$f$$ be a bounded measurable function on a set of finite measure $$E$$. For a measurable subset $$A$$ of $$E$$, show that $$\int_{A} f = \int_{E} f\cdot \chi_{A}.$$

My proof was:

Let $$f$$ be a bounded measurable function on a set of finite measure $$E.$$ Then by Theorem 4 on page 74, $$f$$ is integrable over $$E.$$

Now, by The definition on page 73, $$f$$ is Lebesgue integrable over $$A$$ (because $$A$$ is a measurable subset of $$E$$ by the assumption of the problem and hence has finite measure) if and only if the following holds: \begin{align*}\int_A f &= \sup\{\int_A\varphi : \varphi \text{ is simple and } \varphi \leq f\} \\ &= \inf\{\int_A\psi: \psi \text{ is simple and } f \leq \psi\}.\end{align*}

Also, $$f\cdot \chi_A$$ is integrable if and only if the the following holds: \begin{align*}\int_E f \cdot \chi_A &= \sup\{\int_E\varphi : \varphi \text{ is simple and } \varphi \leq f\cdot \chi_A\} \\ &= \inf\{\int_E\psi: \psi \text{ is simple and } f \cdot \chi_A \leq \psi\}.\end{align*}

Now, since $$\int_A f=\inf\{ \int_A \psi: \psi \text{ is simple and } \psi\geq f \text{ on }A\}$$ and $$\int_E f\cdot\chi_A = \inf \{ \int_E \phi: \phi \text{ is simple and }\phi\geq f\cdot\chi_A \text{ on }E \}.$$

For any given simple function $$\psi$$ such that $$\psi\geq f$$ on $$A,$$ we can extend it so that $$\psi=0$$ on $$E\setminus A$$ and this extension is still a simple function.

Therefore, for any $$x\in E,$$ $$(f \cdot \chi_A)(x) = \begin{cases} f(x) & \text{ if } x\in A \\ 0 & \text{ if } x\in E\setminus A \end{cases} \leq \begin{cases} \psi(x) & \text{ if }x\in A \\ 0 & \text{ if }x\in E\setminus A \end{cases} = \hat{\psi}(x).$$

Now, if $$\psi \geq f$$ on $$A$$, then $$\psi \cdot \chi_A \geq f \cdot \chi_A$$ on $$E$$ by monotonicity of integration proposition 2 or Theorem 5 and because for simple functions we have $$\int_A \psi = \int_E \psi \cdot \chi_A$$.

Thus,

$$\int_A \psi = \int_E \psi \cdot \chi_A \geq \inf_{\hat{\psi} \geq f \cdot \chi_A} \int_E\hat{\psi} = \int_E f \cdot \chi_A.$$

Taking the infimum of the LHS, we obtain

$$\int_A f = \inf_{\psi \geq f} \int_A \psi \geq \int_E f \cdot \chi_A.$$

Hence, $$\int_A f \geq \int_E f\cdot\chi_A$$.

Now,to show that $$\int_A f \leq \int_E f \cdot \chi_A$$, let $$\phi$$ be a simple function such that $$\phi \leq f$$ on $$A$$. It follows that $$\phi \cdot \chi_A \leq f \cdot \chi_A$$ on $$E$$ and

$$\int_A \phi = \int_E \phi \cdot \chi_A \leq \sup_{\hat{\phi} \leq f \cdot \chi_A}\int_E \hat{\phi} = \int_E f \cdot \chi_A.$$

Taking the supremum of the LHS, we obtain

$$\int_A f = \sup_{\phi \leq f} \int_A \phi \leq \int_E f \cdot \chi_A.$$

1-Why is $$f$$ measurable on $$A$$?

2-Why is $$f\cdot \chi_{A}$$ measurable?

3- Prove that for simple functions we have $$\int_{A} \psi = \int_{E} \psi \cdot \chi_{A}$$?

Note: we are not allowed to use any material from the book after page 79.

1. Let $$M \subseteq \mathbb{R}$$ be Borel measurable. Since $$f$$ is a measurable function, the preimage $$f^{-1}(M)$$ is measurable. Since $$A$$ is measurable, $$f^{-1}(M) \cap A$$ is measurable.
2. As before, let $$M \subseteq \mathbb{R}$$ be Borel measurable. Then $$(f\cdot \chi_A)^{-1}(M) = \begin{cases}f^{-1}(M) \cap A, & \text{ if } 0\not\in M \\ (f^{-1}(M) \cap A) \cup (E\setminus A), & \text{ if } 0 \in M \end{cases}$$ which is measurable in either case since $$f$$ is a measurable function and $$A \subseteq E$$ is measurable.
3. Let $$N \subseteq E$$ be measurable. Then $$\int_A \chi_N = |N\cap A| = \int_E\chi_{N\cap A} = \int_E\chi_N\cdot \chi_A$$ shows that $$\int_A\psi = \int_E \psi\cdot \chi_A$$ is true when $$\psi = \chi_N.$$ By linearity of the integral, it is also true when $$\psi$$ is a simple function.
• You should probably clarify that you mean that $M$ is Borel measurable. For a Lebesgue measurable function, the inverse image of a Lebesgue measurable set need not be Lebesgue measurable (see math.stackexchange.com/questions/479441/…). – cmk Oct 16 at 22:44
• @cmk In the present context, $M$ must be measurable in the codomain of $f,$ which certainly means $M$ is implicitly Borel measurable for Lebesgue measurable $f.$ In any case, thanks for pointing out the possible confusion someone might have when learning this topic. – Brian Moehring Oct 16 at 23:09
• For 3, what do you mean by the absolute value sign in $|N \cap A|$? – Emptymind Oct 17 at 12:04
• @Intuition The measure of the the set $X$ is sometimes written $|X|.$ In particular, in this type of case where have no "name" for the measure (common names being $\nu,$ $\mu$ or $m$), $|N \cap A|$ allows us to denote the measure of $N \cap A$ without introducing another symbol. – Brian Moehring Oct 18 at 3:43

$$1$$. If $$V$$ is an arbitrary open set in $$\mathbb{R}$$ and $$f|_A$$ denotes the restriction of $$f$$ to $$A$$, then $$(f|_{A})^{-1}(V)=f^{-1}(V)\cap A.$$ A real-valued function is Lebesgue measurable if and only if its inverse image of an open set is measurable. Since $$f$$ is measurable, so is $$f^{-1}(V),$$ and $$A$$ is measurable by assumption. So, their intersection is measurable.

$$2$$. For any measurable functions, $$f$$ and $$g$$, I claim that $$fg$$ is measurable. First, note that $$fg=\frac{(f+g)^2-f^2-g^2}{2},$$ so it will suffice to show that if $$h$$ is measurable, then so is $$h^2$$ (since the sum of measurable functions is measurable and so is a measurable function times a constant, both of which I assume you know; if not, they follow from the composition property that I'll cite below). Note that this is a composition of $$h$$, which is measurable, and $$x^2$$, which is continuous, so their composition will be measurable. This is because if $$u$$ is continuous and $$v$$ is measurable, then $$u\circ v$$ is measurable, too; this follows from $$(u\circ v)^{-1}(V)=v^{-1}\circ u^{-1}(V)$$, since $$u^{-1}(V)$$ is open for $$V$$ open by continuity and $$v$$ is measurable, so the inverse image of an open set is measurable. If you don't like using a result like this, then you can instead check measurability on $$(a,\infty),$$ for any $$a$$. The inverse image for $$a<0$$ is everything, and for $$a\geq 0$$ is $$\{x: h^2(x)>a\}=\{x:h(x)>\sqrt{a}\}\cup\{x:-h(x)>\sqrt{a}\},$$ which is clearly measurable.

In any case, $$f$$ and $$\chi_A$$ are measurable, so is their product. You can do this more explicitly since you're working with something like a characteristic function, but we can pretty easily work in more generality, as shown.

$$3$$. Let $$\psi(x)=\sum\limits_{j=1}^nc_j\chi_{A_j}(x),$$ where $$A_j$$ are disjoint and measurable. Then, $$\int\limits_A \psi=\sum\limits_{j=1}^nc_jm(A_j\cap A),$$ and \begin{align*}\int\limits_E \psi\chi_A=\int\limits_E \sum\limits_{j=1}^nc_j\chi_{A_j}\chi_A&=\int\limits_E \sum\limits_{j=1}^nc_j\chi_{A_j\cap A}=\sum\limits_{j=1}^n c_j\int\limits_E \chi_{A_j\cap A}\\ &=\sum\limits_{j=1}^nc_jm(A_j\cap A). \end{align*} So, they indeed match. Here, I used the definition of the integral of a simple function, properties of characteristic functions (what their product looks like), and linearity of the integral.

• @Intuition it suffices to check measurability on open sets (that is, to show that the inverse image of any open set is measurable). – cmk Oct 16 at 22:17
• This is because open sets generate the Borel sets. – cmk Oct 16 at 22:23
• @Intuition the question was to show that $f$ is measurable on $A$. That means that $f|_A:A\rightarrow\mathbb{R}$ is a Lebesgue measurable function. To show this, we need to show that $(f|_A)^{-1}(V)$ is Lebesgue measurable for every open set $V\subset\mathbb{R}.$ I don't need to have an open $V$ or anything. I need to show that the statement is true for an arbitrary open $V$. – cmk Oct 16 at 22:54
• @Intuition whoops, typo. Fixed now. Thanks! – cmk Oct 17 at 0:46
• @Intuition for my representation of $\psi,$ yes, I'm using pages $71-72$. The reason that I'm intersecting with $A$ when I'm taking the measure is because the $A_j$'s are measurable subsets of $E$, and I'm integrating over just $A$ (I should've noted that $A_j\subset E$ for each $j$). It's pretty easy to see that $\int_A\chi_{A_j}=m(A\cap A_j)$, then we just use linearity. Remember, $\chi_{A_j}$ is only non-zero on $A_j$, and I'm integrating over $A$. So, when we calculate the integral, we're only going to pick up the measure of the part of $A_j$ that's contained in $A$. – cmk Oct 17 at 13:15