Having trouble justifying order of integral and sum

Question

I'm trying to prove that $\mu$ is a measure on $\mathcal{M}$, where $\mu(E) = \int_{E}f$, $f: \mathbb{R} \to [0, \infty]$ is a Lebesgue measurable funciton, and $\mathcal{M}$ consists of all the Lebesgue measurable subsets of $\mathbb{R}$.

Right now, I'm trying to show countable additivity: $\mu(\cup_{k=1}^{\infty}E_{k}) = \cup_{k=1}^{\infty}\mu(E_{k})$, where the $E_{k}$ are disjoint.

So far, I have that $\mu(\cup_{k=1}^{\infty}E_{k}) = \int_{\cup_{k=1}^{\infty}E_{k}}f = \int f 1_{\cup_{k=1}^{\infty}E_{k}}$ (where $1_{\cup_{k=1}^{\infty}E_{k}}$ is an indicator function) $= \int \sum_{k=1}^{\infty}f 1_{E_{k}}$, and on the next step, I need to switch the $\int$ and the $\sum$. I would like to use Monotone Convergence Theorem to get the job done, but $f 1_{E_{k}}$, while nonnegative, is not necessarily increasing (or is it?).

So, how do I get this to look like something that is monotone increasing so that I can apply the theorem I need?

Answer 1

Hint: Define \begin{align}B_1&:=E_1\\B_2&:=E_1\cup E_2=B_1\cup E_2\\ B_3&:=E_1\cup E_2 \cup E_3=B_2\cup E_3\\\ldots& \qquad \ldots\\B_{k+1}&:=B_k\cup E_{k+1}\end{align} Then $(f\mathsf1_{B_k})_{k\in\mathbb N}$ is increasing and $\cup_{k=1}^{\infty}B_k=\cup_{k=1}^{\infty}E_k$.

Answer 2

To apply the Monotone Convergence Theorem, note that $\{g_{n}\}_{n=1}^{\infty}$ is an increasing sequence of non-negative measurable functions on $\mathbb{R}$, where $g_{n} := \sum_{k=1}^{n}f\cdot 1_{E_{k}}$. Thus, \begin{align} \sum_{k=1}^{\infty}\int f\cdot 1_{E_{k}} = \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\int f\cdot 1_{E_{k}} = \lim_{n\rightarrow\infty}\int \sum_{k=1}^{n}f\cdot 1_{E_{k}} & = \lim_{n\rightarrow\infty}\int g_{n} \\\\ (MCT) & = \int \lim_{n\rightarrow\infty}g_{n} \\\\ & = \int \sum_{k=1}^{\infty} f\cdot 1_{E_{k}}. \end{align}