The following is part of a problem from one of our previous sample exams:
Let $E$ be a measurable subset of $\mathbb{R}$. Suppose $f$ : $E → \mathbb{R}$ is a measurable function, $f_n$ : $E → \mathbb{R}$ is an increasing sequence of measurable, real-valued functions such that $f_k→f$ a.e. For a real number $a$, define $M_f(a):=m\{x\in E:f(x)>a\}$, where $m\{\cdot\}$ denotes the Lebesgue measure. Then $M_{f_k}\leq M_{f_{k+1}}$ and $M_{f_k}→M_f$.
The first (inequality) part is obvious, but I'm have trouble proving the second (convergence) part though it looks straightforward and intuitive. I wrote out the set in which $f(x)>a$ and $f_k(x)\leq a$ and tried to rewrite with things like $a-\frac{1}{k}$, but could not relate it to the measure.
Any hint or help will be much appreciated!
I'll denote the Lebesgue measure by $m$ since you are using it.
Notice that:
$$M_{f_{k}}(a)=\int_{\Bbb{R}}\mathbf{1}_{\{f_{k}>a\}}\,dm$$
Now see that $\mathbf{1}_{\{f_{k}>a\}}\leq \mathbf{1}_{\{f_{k+1}>a\}} $ and it is an increasing sequence of non-negative functions and $\mathbf{1}_{\{f_{k}>a\}}\to \mathbf{1}_{\{f>a\}}$ almost everywhere.
To see the above, let $A$ be a measure $0$ set, outside of which you have pointwise convergence of $f_{k}\to f$.
Then for $x\notin A$, you have either $f(x)>a$ or $f(x)\leq a$. So $\mathbf{1}_{\{f>a\}}(x)=1 \,\text{or}\, 0$ . If $f(x)>a$, then for all large $k$, $f_{k}(x)>a$ and hence $\mathbf{1}_{\{f_{k}>a\}}(x)=1\,,$ for all large $k$ as $f_{k}(x)\to f(x)$. If
If $f(x)\leq a$, then $\mathbf{1}_{\{f>a\}}(x)=0$ .
Then as $f_{k}(x)$ is an increasing sequence , then $f_{k}(x)\leq f(x)\leq a\,,\forall k\in\Bbb{N}$ and thus $\mathbf{1}_{\{f_{k}>a\}}(x)=0$ for all $k$ .
Thus you have almost everywhere convergence of $\mathbf{1}_{\{f_{k}>a\}}\to \mathbf{1}_{\{f>a\}}$.
Thus By Monotone Convergence Theorem you have:-
$$\lim_{k\to\infty}M_{f_{k}}(a)=\lim_{k\to\infty}\int_{\Bbb{R}}\mathbf{1}_{\{f_{k}>a\}}\,dm=\int_{\Bbb{R}}\lim_{k\to\infty}\mathbf{1}_{\{f_{k}>a\}}\,dm=\\\int_{\Bbb{R}}\mathbf{1}_{\{f>a\}}\,dm=m(\{x\in E:f(x)>a\})=M_{f}(a)$$