Given $f$, $f_{1}$, $f_{2,...}$ are monotone functions on $(a, b) \subset \mathbb{R}$ and $f_{n}$ converges to $f$ in the Lebesgue measure on$(a, b)$. Prove that $f_{n}(x) \to f(x)$ at every point $x$ where $f$is continuous at $x$.
This is my proof
For $\epsilon > 0$, consider the set$$B = \cup_{n=1}^{\infty}\cap_{n=1}^{\infty}\{x \in (a, b) : |f_k(x) - f(x)| < \epsilon \}.$$We prove that if $f$ is continuous at $x_0$, then $x_0 \in B$ and therefore the required result is obtained. Alternatively, we prove that if $x_0 \notin B$, then $f$ is not continuous at $x_0$. Suppose $x_0 \notin B$. Then, there exists an increasing sequence $k_n \to \infty$ such that for every $n$ we have$$|f_{k_n}(x_0) - f(x_0)| \geq \epsilon.$$On the other hand, the sequence $\{f_{k_n}\}$ converges to $f$ in measure, so it can be assumed to converge almost everywhere to $f$ on $(a, b)$. Therefore, there exists a set $D \subset (a, b)$ with $\mu_{1}(D) = 0$, and for every $x \in (a, b) \setminus D$, we have $\lim\limits_{n \to \infty} f_{k_n}(x) = f(x)$. Thus, a sequence $\{x_m\} \subset (a, b)$ can be found such that $\lim\limits_{m \to \infty} x_m = x_0$ (by the Weierstrass theorem). Assume that for large enough $n$, $f_{k_n}$ is continuous at $x_0$. Then, for sufficiently large $m$, we have\begin{align*}|f(x_m) - f(x_0)| &= |f(x_m) - f_{k_n}(x_m) + f_{k_n}(x_m) - f_{k_n}(x_0) + f_{k_n}(x_0) - f(x_0)| \\&\geq |f_{k_n}(x_0) - f(x_0)| - |f_{k_n}(x_m) - f_{k_n}(x_0)| - |f(x_m) - f_{k_n}(x_m)| \\&\geq \tfrac{\epsilon}{3}.\end{align*}Thus, $f$ is discontinuous at $x_0$.
Is this proof correct? And what is the role of monotonicity? It seems that I did not use monotonicity in this solution.