# Show that $E=\bigcup_{k=1}^{\infty}E_k$, where for each index $k, E_k$ is measurable, and $(f_n)$ converges uniformly to $f$

Let $(f_n)$ be a sequence of measurable functions on $E$ that converges to the real-valued $f$ pointwise on $E$. Show that $E=\bigcup_{k=1}^{\infty}E_k$, where for each index $k, E_k$ is measurable, and $(f_n)$ converges uniformly to $f$ on each $E_k$ if $k>1$, and $m(E_1)=0.$

My solution: Let $f_n$ converge pointwise to $f$. For any $\frac{1}{k}$ there exists $E_k$ measurable such that $m(E-E_k) < \frac{1}{k}$ and $f_n \to f$ uniformly on $E_k$ (By Egoroff's Theorem).

We have $E - \bigcup\limits_{k=2}^{\infty} E_k \subset (E - E_k)$, therefore $m(E - \bigcup\limits_{k=2}^{\infty} E_k)=0$. Consider $E_1=E -\bigcup\limits_{k=2}^{\infty} E_k$, then $E=\bigcup\limits_{k=1}^{\infty} E_k$ and $m(E_1) = 0$.

Anyone can check my solution? Thanks

• Your solution seems to correct :) – Mike Jul 14 '16 at 20:55
• I don't think you can apply Egoroff's Theorem because you don't know that $m(E_k) < \infty$ – David Apr 6 '17 at 15:22

In the comment, David mentioned that Egoroff's Theorem does not apply when $m (E) = \infty$. In fact, the question does not assume that $m (E) < \infty$. We need to take it into consideration. I think the proof is correct in the original post when $m (E)$ is finite. So one way to resolve the infinite measure of $E$ is use \begin{align*} E = \cup_{n \in \mathbb{Z}} E \cap [n, n+1] = \cup_{n \in \mathbb{Z}} E_n \end{align*} where $E_n = E \cap [n ,n+1]$. Then, for each $n$, we apply the proof in the original post and get a countble set $\{ E_{n,i}\}_{i=2}^\infty$ where the convergence is uniform, and $E_{n,1}$ with $m(E_{n,1}) = 0$. Let \begin{align*} \{ E_k \}_{k=2}^\infty = \cup_{n \in \mathbb{Z}} \{ E_{n,i} \}_{i=2}^\infty \end{align*} by re-ordering the index. And let \begin{align*} E_1 = \cup_{n \in \mathbb{Z} } E_{n,1}. \end{align*} We have \begin{align*} \mu (E_1) \leq \sum_{n \in \mathbb{Z} } \mu (E_{n,1}) = 0. \end{align*} The proof is finished.