$\lim_{n\to\infty}\int_X f_n d\mu=2023$ and measures

Question

Let $(X,\mathcal{A},\mu)$ be a measure space and let $\{f_n\}_{n\geq1}$ be a sequence in $\mathcal{M}_{\bar{\mathbb{R}}}^+(\mathcal{A})$. Assume that $f_n\to f$ for $n\to \infty$ and that

$$\lim_{n\to\infty}\int_X f_n d\mu=2023$$

Argue that $f\in\mathcal{M}_{\bar{\mathbb{R}}}^+(\mathcal{A})$ and that $\int_X f d\mu\leq 2023$. Argue that if $f_1\leq f_2\leq\ldots.$ then $\int_X f d\mu=2023$

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$\mathcal{M}_{\bar{\mathbb{R}}}^+(\mathcal{A})$ is for the families of (extended) real-valued positive measurable functions.

I am thinking that since the sequence is in $\mathcal{M}_{\bar{\mathbb{R}}}^+(\mathcal{A})$ and $f_n\to f$ for $n\to\infty $ then $f(=\sup f_n?)$ must also be in $\mathcal{M}_{\bar{\mathbb{R}}}^+(\mathcal{A})$. I have been using this theorem but unsure on how to apply it exactly:

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Theorem: Let $(X, \mathcal{A})$ be a measurable space. If $u_n : X \to \bar{\mathbb{R}}$, $n \in \mathbb{N}$, are measurable functions, then so are $$ \sup_{n \in \mathbb{N}} u_n, \quad \inf_{n \in \mathbb{N}} u_n, \quad \limsup_{n \to \infty} u_n, \quad \text{and} \quad \liminf_{n \to \infty} u_n, $$ and, whenever it exists in $\bar{\mathbb{R}}$, $\lim_{n \to \infty} u_n$.

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For the two other claims regarding the integral I am not sure whatsoever. It seems awful lot like a set up for Beppo Levi but I have not applied the theorem before so I am unsure on both the approach and if I am allowed to.

Thank you so much.

Answer 1

• The hypothesis that $f_n\to f$ as $n\to\infty $ does NOT imply $f=\sup f_n$. Indeed, we have $\lim\limits_{n\to\infty}f_n=f=\limsup\limits_{n\to\infty}f_n=\liminf\limits_{n\to\infty}f_n$, and thus $f$ is measurable and non-negative provided that $f_n\in\mathcal{M}_{\bar{\mathbb{R}}}^+(\mathcal{A})$ for each $n\in\mathbb N_{\geq1}$.

• By Fatou's lemma, we have $$\int_X \liminf\limits_{n\to\infty}f_n\,d\mu\leq \liminf_{n\to\infty}\int_X f_n\,d\mu.$$ Combining the above with $f=\liminf\limits_{n\to\infty}f_n$ and $\lim\limits_{n\to\infty}\int_X f_n d\mu=2023$ gives that $\int_Xf\,d\mu\leq 2023$.

• If $\{f_n\}_{n\geq1}$ is moreover a non-decreasing sequence, then Beppo Levi's lemma implies that $$\int_X f\,d\mu =\lim\limits_{n\to\infty}\int_X f_n d\mu=2023.$$