If $f_n \to f$ pointwise and $f_n$'s are measurable, then is $f$ measurable? Take a sequence of functions $f_n$ which are measurable i.e. $f_n \in L^+$. And $f_n \to f$ pointwise. Then is this imply $f$ is measurable? If not, why? Thanks.
 A: Hint: Given $a\in\mathbb{R}$, $\{x:f(x)>a\}=\cup_{n=1}^\infty\cap_{k=n}^\infty\{x:f_k(x)>a\}$.
A: This is true. Let $f_{n}\rightarrow f$ pointwise. We proceed in three steps: showing $\sup_{n}f_{n}$ is measurable, $\limsup_{n}f_{n}$ is measurable, the same for $\inf$ and $\liminf$, and then combine these to get that $\lim f_{n}$ is measurable. I'll give you proofs for $\sup$ and $\limsup$, it should be simple to modify them for $\inf$ and $\liminf$.
$\sup_{n}f_{n}(x)>a$ iff $f_{n}(x)>a$ for some $n$, and so 
$$\left\{x\in\Bbb{R} \ | \ \sup_{n}f_{n}(x)>a\right\}=\bigcup_{n=1}^{\infty}\{x\in\Bbb{R} \ | \ f_{n}(x)>a\}.$$
Each set in the union is measurable by assumption, so the union is measurable as measurable sets form a sigma algebra. Hence $\sup_{n}f_{n}$ is measurable.
Now, write $g_{n}=\sup_{k\geq n}f_{k}$. Then $g_{n}$ is measurable for each $n$ by the above. For each $x$, $g_{n}$ is a decreasing sequence, so we have
$$\limsup_{n\rightarrow\infty}f_{n}(x)=\lim_{n\rightarrow\infty}g_{n}(x)=\inf_{n}g_{n}(x),$$
and so the first part gives us that $\limsup_{n}f_{n}$ is mesaurable.
Finally, $f_{n}\rightarrow f$ pointwise iff $\limsup_{n\rightarrow\infty}f_{n}=\liminf_{n\rightarrow\infty}f_{n}=f$, so together with the (almost identical) proof of the $\inf$ case, we have the result.
A: Yes. In Rudin(Real and Complex analysis), he proves $\limsup f_n$ and $\liminf f_n$ are measurable. In your case, the limit itself exists pointwise and so is equal to either of these.
