The problem is stated below.

If $\mu(E_n) < \infty$ for $n \in \mathbb{N}$, and $\chi_{E_n} \to f$ in $L^1$, then $f$ is a.e. equal to the characteristic function of a measurable set.

If $\chi_{E_n} \to f$ in $L^1$, then the sequence of characteristic functions also converges to $f$ in measure, thus there exists a subsequence of measurable functions which converges to $f$ a.e. (since characteristic functions are measurable). The a.e. limit of measurable functions is measurable, so $f$ is measurable.

Since each characteristic function in our a.e. convergent subsequence is either 0 or 1 at any particular $x \in X$, the limit, $f(x)$, is either 0 or 1. Since $\{1\}$ and $\{0\}$ are Borel sets, and $f$ is measurable, $f^{-1}(\{1\})$ and $f^{-1}(\{0\})$ are contained in our sigma algebra. So define \begin{align*} F \doteq \{x \mid f(x) = 1\} \end{align*} $f$ is thus the characteristic function over the measurable set $F$.

I think this is incomplete because I never used the assumption $\mu(E_n) < \infty$ in my proof, and also I'm not sure if I can just say the $\mu$ a.e. limit of measurable functions is measurable unless I prove that $\mu$ is complete.

  • $\begingroup$ If $\mu(E)=\infty$, then $1_E\notin L^1$. (After all, what if $\{f_n\}$ were s.t. $\int |f_n-f|\to 0$ but $f\notin L^1$?) Also, Folland's proof that you can find $f_{n_k}\to f$ $\mu$-a.e. should be independent of the completeness of the measure. $\endgroup$ – Nobody Jan 4 '17 at 8:16
  • $\begingroup$ And you don't need that $\mu$ is complete for the mesaurability of the limit… to see this make clear that $\limsup f_n$ and $\liminf f_n$ are measurable due to unions and intersections of measurable sets and $$\{\lim f_n \text{ exists }\} = \{ \limsup f_n = \liminf f_n \}$$ holds. Hence $\lim f_n$ is measurable. So I guess your proof is fine. $\endgroup$ – Gono Jan 4 '17 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.