Convergence almost uniformly Let $(X,\Sigma,\mu)$ a measure space. Let $E_n\in\Sigma$ such that $\mu(E_n)>0$ and we suppose that $f_n:=a_n\chi_{E_{n}}$, where $a_n>0$, converges almost uniformly to $0$. Then I want to prove that $a_n\to 0$  or $\mu\left(\bigcup_{j\ge N}E_j\right)\to 0$.
Additional hypothesis: $a_n\to 0$ or $a_n\ge C$ for some $C>0$ and every $n\in\mathbb{N}$ (I don't know if this is neccesary, but we have it anyway).
I don't know if this is helpless, but I tried this:
For each $n\in\mathbb{N}$ there exists $K_n\in\Sigma$ such that $\mu(K_n)<\dfrac{1}{2^n}$ and $f_n\rightrightarrows 0$ in $K_n^c$ (the complement of $K_n$).
Case 1. There is $M\in\mathbb{N}$ such that $m\ge M$ implies $K_m^c\cap E_m=\emptyset$. So if $m\ge M$ we have $E_m\subseteq K_m$ and $$\mu\left(\bigcup_{j\ge m}E_j\right)\le \sum_{j\ge m}\dfrac{1}{2^j}\to 0.$$
Case 2. For each $M\in\mathbb{N}$ there exists $m\ge M$ such that $K_m^c\cap E_m\neq\emptyset.$
I don't know what to do in this case, but what I wanted was something like this: by the uniform convergence in some $K_m^c$ and by taking $x_m\in K_m^c\cap E_m$ we will have $|f_m(x_m)|=|a_m\chi_{E_m}(x_m)|=|a_m|<\epsilon$ for a fixed $\epsilon >0$, but I can't do that yet. Is this useful anyway?
Thanks for any answer and sorry for my english.
 A: Do you think that you could try going directly by definition?
By definition, for any $\gamma > 0,\;\exists A_\gamma\subset X$ with $\mu(A_\gamma) < \gamma$ such that $\forall\varepsilon > 0,\;\exists N > 0$ such that $|f_n| = |a_n\chi_{E_n}| < \varepsilon,\; \forall n\geq N,\;\forall x\in X\backslash A_\gamma$. (I just wrote what it means, to converge uniformly on $X\backslash A_\gamma$, which is the definition of almost uniform convergence).
Now, you play a little with the $\chi$:
For any $\gamma > 0,\;\exists A_\gamma\subset X$ with $\mu(A_\gamma) < \gamma$ such that $\forall\varepsilon > 0,\;\exists N > 0$ such that $|f_n| = |a_n\chi_{\color{blue}{E_n\backslash A_\gamma}}| < \varepsilon,\; \forall n\geq N,\;\forall \color{blue}{x\in X}$.
Let us suppose that $a_n\not\to 0$. Then there exists an epsilon small enough such that $\forall N > 0, \exists n \geq N: a_n > \varepsilon$. But nevertheless, $|a_n\chi_{E_n\backslash A_\gamma}| < \epsilon$ for all x\in X. This means that $|a_n| < \epsilon$ for all $x\in E_n\backslash A_\gamma$. Which means that $E_n\backslash A_\gamma = \emptyset$ for this given $n$. Which means that $E_n\subseteq A_\gamma$ and $\mu(E_n)\leq \mu(A_\gamma) < \gamma$. By selecting a sequence of $\gamma_i$ such that $\gamma_{j+1} < \gamma_j$, you can make $\mu(E_n)$ arbitrarily small for this particular $n$.
Now, you can use the fact that if $a_n\not\to 0$, then $a_n > C$ where $C$ is some constant. And you're done.
Here's an example: Say, you have a sequence $a_n = 1/n$, but when $n \in \{10, 20, 30, ...\}$, you have $a_n = 1$. This sequence doesn't converge, because every tenth element is equal to $1$. Now, choose $E_n = [0, 1]$, except when $n \in \{10, 20, 30, ...\}$, you put $E_n = [0, 1/n]$. Choose any $\gamma$. You can always find $A_\gamma = [0, a]$ such that $\mu(A_\gamma) < \gamma$, and by removing it, you have that starting with some $N$, these "small" sets $E_n\subset A_\gamma$ and so your function is zero in $X\backslash A_\gamma$ for all the $n$ that are divisible by $10$.
As for other $n$, they converge to $0$ uniformly. So, we constructed a sequence of $a_n$ and a sequence of corresponding $E_n$ such that $a_n\not\to 0$ and $\mu(E_n)\not\to 0$, but your $|f_n|\to 0$ almost uniformly. The key here is that there is no constant $C$ such that $a_n > C$.
