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Prove that if $f_j : A \to \Bbb R, j \in \Bbb N$ are measurable, then $\limsup_{j\to \infty} f_j$ is measruable.

First I tried to show that $\sup_j f_j$ is measurable. Now according to the definition $\sup_j f_j$ is measurable if $\forall a \in \Bbb R$ we have that $\{x \in A \mid \sup_j f_j(x) > a \}$. This means that for some $j \in \Bbb N$ we have that $f_j(x) > a$ right? So $\{x \in A \mid \sup_j f_j(x) > a \} = \sup_{j}f_j^{-1}[(a, \infty]] = \bigcup_{j} f_j^{-1}[(a, \infty])$ and these sets are measurable since $(a, \infty]$ is a Borel set and the countable union of measurable sets is measurable?

Now by definition $\limsup_{j \to \infty} f_j = \lim_{j \to \infty} (\sup_{k\ge j} f_k)$ and now I know that $\sup_{k\ge j} f_k$ is measurable, but how about the limit? For sets the $\limsup_j A_j = \bigcap_{j} \bigcup_{k\ge j} A_k$, but I'm not working with sets here?

UniQGat
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1 Answers1

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Claim 1: $f:=\sup_{j}f_{j}$ is measurable.

Let $\mathcal{C}=\{[-\infty,a]\mid a\in\mathbb{R}\}$. Note that $\sigma(\mathcal{C})=\mathcal{B}([-\infty,\infty])$. To show that $f$ is measurable, it suffices that $f^{-1}(B)$ is measurable for each $B\in\mathcal{C}$. Let $B\in\mathcal{C}$, then $B=[-\infty,a]$ for some $a\in\mathbb{R}$. We assert that $f^{-1}(B)=\cap_{j}f_{j}^{-1}(B)$. For, let $x\in LHS$, then $f(x)\leq a$. It follows that $f_{j}(x)\leq f(x)\leq a$ for each $j$ and hence $x\in RHS$. Conversely, suppose that $x\in RHS$, then $f_{j}(x)\leq a$ for each $j$. This shows that $a$ is an upper bound of $\{f_{j}(x)\mid j\in\mathbb{N}\}$ and hence $\sup\{f_{j}(x)\mid j\in\mathbb{N}\}\leq a$. That is, $f(x)\leq a$ and hence $x\in LHS$. Now it is clear that $f^{-1}(B)$ is measurable.

For your case, for each $j\in\mathbb{N}$, let $g_{j}=\sup_{k\geq j}f_{k}$. By claim 1, $g_{1},g_{2},\ldots$ are measurable. Note that $g_{1}\geq g_{2}\geq\ldots$, so $\lim_{j}g_{j}=\inf_{j}g_{j}$, which is measurable. (Measurability of $\inf_j g_j$ can be proven similarly.) However, $\lim_{j}g_{j}=\limsup_{j}f_{j}$.

Danny Pak-Keung Chan
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