# Prove $\lim\limits_{k\to \infty} f_k(x) =f(x)$ a.e. $x \in \mathbb{R}^d$.

Let $$f$$ and $$\lbrace f_k \rbrace_{k=1}^\infty$$ be integrable function on $$\mathbb{R}^d$$. Suppose for any measurable set $$E \subset \mathbb{R}^d$$ there holds: \begin{align*}\displaystyle\int_E f_k(x)dx < \displaystyle\int_E f_{k+1}(x)dx, \quad k\in\mathbb{N} \end{align*} and \begin{align*}\lim\limits_{k\to \infty}\displaystyle\int_Ef_k(x)dx = \displaystyle\int_Ef(x)dx. \end{align*} Show that $$\lim\limits_{k\to \infty} f_k(x) =f(x)$$ a.e. $$x \in \mathbb{R}^d$$.

And this is my attempt, let $$F = \lbrace x \in \mathbb{R}^d: f_k(x) \text{ doesn't converge to } f(x) \text{ as } k \to \infty \rbrace$$. Then \begin{align*} F=\displaystyle \bigcup\limits_{m=1}^\infty \bigcap\limits_{j=1}^\infty\bigcup\limits_{k=j}^\infty \left\lbrace x \in \mathbb{R}^d: |f_k(x)-f(x)| >\dfrac{1}{m} \right\rbrace \end{align*} So $$F$$ is measurable and there exists $$m \in \mathbb{N}$$ such that \begin{align*} |f_k(x) -f(x)| >\dfrac{1}{m} \\ \displaystyle \int_F |f_k(x) -f(x)|dx > \dfrac{1}{m} m(F) \end{align*} I want to prove that as $$k \to \infty$$, $$m(F)=0$$, but I don't know how. Can you give me any idea? Any idea is highly appreciated.

Hints: $$\int_E f_k(x)dx < \int_E f_{k+1}(x)dx$$ for all measurable set $$E$$ implies that $$f_k \leq f_{k+1}$$ a.e.. This implies that $$g(x)=\lim f_k(x)$$ exists a.e.. By Monotone Convergence Theorem we get $$\int_E g(x)dx=\lim_{n \to \infty} \int_E f_n(x)dx=\int_E f(x)dx$$. Since this holds for all $$E$$ we see that $$f=g$$ a.e.