Prove that if $f$ is nonnegative, measurable and $E_k \nearrow E$, then $\lim_k \int_{E_k} f = \int_E f$.

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)