# Need a little help with this {f_n} converging to f(x) proof

## Homework Statement

Suppose that $$f_n \stackrel{P}{\rightarrow} f$$ on $$S$$. Given $$\epsilon >0$$ define $$E_{\epsilon,n}=\{x\in S \mid |f_n(x)-f(x)|<\epsilon\}.$$
Prove that $$\forall \epsilon >0$$ $$\cup_{n=1}^\infty E_{\epsilon,n}=S$$.

## Homework Equations

Know about measurability, pointwise convergence etc.

## The Attempt at a Solution

I came up with this proof, but I'm not entirely sure it's right. It's also not very formalized..but besides that it almost seems like it's missing something:

Assume {f_n} -> f pointwise on S, and given epsilon>0 define E as above. Note that E = $$\bigcap_{n=1}^\infty \{x \mid |f_n(x) - f(x)| < \epsilon\}.$$
Since each f_n is measurable and f_n -> f pointwise, then we know f_n-f is measurable for all n in N, and so $$\{x \mid f_n(x) - f(x)| < \epsilon\}$$ is measurable for all n in N, thus E_{n,\epsilon} is a countable intersection of measurable sets and therefore measurable.
Note that E_{n,epsilon} is contained in E_{n+1,epsilon} for all n in N. Note also that since f_n -> f pointwise for all x in S, f_n(x) -> f(x) as n->\infty. So for all x in S there exists a k in N such that whenever N\geq k |f_N(x) - f(x)| < \epsilon. Hence for all x in S, x is in E_{n,\epsilon} for some n in N. Thus $$\bigcup_{n=1}^\infty E_{n, \epsilon} = S,$$ as desired.

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