Convergence a.e. of this characteristic function let $f \geq 0$ be measurable on $E \subset \mathbb{R^n}$.
Now for any $\ n \geq 1$ there is a subset $E_n$ of $E$ with $\ m(E - E_n) < \frac{1}{n}$
I want to show that if I define: $f_n = f \chi_{_{E_n}}$
then $\lim\limits_{n \rightarrow \infty}f \chi_{_{E_n}} = f \chi_{_{E}} = f \ a.e.$
Is this fact obvious by construction and from the givens since measure of the $E_{_n}^{^c}$ gets smaller and converges to $0$
 A: No this claim is incorrect. Let $f\equiv1$, $E=[0,1]$, and $F_n$ a sequence of intervals in $E$ with $|F_n|=1/(2n)$ whose union covers $E$ infinitely-many times. (This is possible because the harmonic series diverges.) Then let $E_n=E\setminus F_n$. Clearly $f\chi_{E_n}$ converges nowhere on $E$.
A: As Funktorality mentioned, it's false. What you really need is for $m(E\setminus E_n)$ to decay sufficiently fast (in particular, in a summable way).
For a.e. convergence to hold we need the following to be true: For a.e. $x\in E$ there exists some $N$ such that $x\in E_n$ for all $n\geq N$. In other words, we need the set on which this doesn't hold -- the set of $x$ such that for all $N$, $x\in E\setminus E_n$ for some $n\geq N$ -- to have measure zero. In set notation, this means
$$m\left(\bigcap_{N=1}^{\infty}\bigcup_{n\geq N}(E\setminus E_n)\right) = m\left(\limsup_{n\to\infty} {(E\setminus E_n)}\right) = 0$$
If the measures of the differences are summable, i.e.
$$\sum_{n=1}^{\infty}m(E\setminus E_n) <\infty$$
Then the desired conclusion is guaranteed by the Borel Cantelli Lemma.
