Pleas check my proof of the following assertion.

Let $\{E_k\}$ be a sequence of measurable sets, and $E_k \nearrow E$. Suppose $f(x)$ is nonnegative and measurable on $E$. Prove

$$\int_E f(x) \, dx = \lim_{k \to \infty} \int_{E_k} f(x) \, dx$$

Proof. First note that

$$\int_{E_k} f(x) \, dx= \int_E f(x)\chi_{E_k}(x) dx$$

and we have (using our hypothesis in the second equality)

$$\lim f(x)\chi_{E_k}(x) = f(x) \lim \chi_{E_k}(x) \stackrel{\tiny{HYP.}}{=} f(x)\chi_E(x) \tag{?}$$

Furthermore, $\left(f(x)\chi_{E_k}(x)\right)$ is monotone increasing, nonnegative, and measurable (product of two measurable functions). Therefore, by the MCT

$$\lim_k \int_{E_k} f(x) \, dx= \lim_k \int_E f(x)\chi_{E_k}(x) \, dx \stackrel{\tiny{MCT}}= \int_E f(x) \, dx$$

So the proof is complete.

I took for granted in $(?)$ that $\lim \chi_{E_k}(x) = \chi_E(x)$. So to prove that we need to show that for all $\epsilon > 0$, and any $x \in E$, there exists and $N$, such that whenever $k > N$ we have

$$ |\chi_E(x) - \chi_{E_k}(x)| < \epsilon$$

But this can be relaxed to almost any $x \in E$ for the MCT. So, let $\epsilon > 0$ and let $x \in E$. The set of $x$ on the boundary of of $E$ has measure zero, since that set has side length zero in at least one dimension (??). If $x$ is in the interior of $E$ we can, by hypothesis, produce an $E_N \subset E$ such that $x \in E_N = E \cap E_N$. Therefore, for all $k > N$ we have $x \in E_k$ which means that $\chi_E(x) = 1$ and $\chi_{E_k}(x) = 1$, which clearly satisfies the inequality above.

Something about that doesn't sit right with me, the (??) part mostly. I was thinking about setting this problem up a different way too.


Not sure if this answers your question, but we can show that if $E_k \uparrow \ E$ then we have that: $$ \mu\left(\bigcup_{k = 1}^{\infty}E_k\right) = \lim_{n \to \infty} \mu\left(\bigcup_{k = 1}^{n}E_k\right) $$ This is done by defining $G_1 = E_1$, $G_2 = E_2 \setminus E_1$, $G_3 = E_3 \setminus E_2$, etc, with $G_ k = E_k \setminus E_{k-1}$. Then, we see that these sets are disjoint, and $\bigcup G_k = E = \bigcup E_k$. By disjoint additivity, we have: $$ \mu(E) = \sum_{k = 1}^\infty \mu(G_k) = \lim_{n \to \infty} \sum_{k = 1}^{n}\mu = (G_k) = \lim_{n \to \infty} \mu \left(\bigcup_{k = 1}^{n}G_k\right) = \lim_{n \to \infty} \mu \left(\bigcup_{k = 1}^{n}E_k\right) $$ as we needed. Having proven the two sets have the same measure, it follows that the indicator function converges accordingly (in an A.E sense)


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