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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.

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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.

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  • $\begingroup$ Oh I see. Thank you so much. I never think of that! $\endgroup$ Nov 4, 2022 at 7:56

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