# Clarification of “Convergence almost everywhere implies convergence in measure”

I have looked over the proofs that show convergence a.e. imply convergence in measure. I understand the proofs, but I do not understand why one must go into such detail.

It seems as though one builds a series out of the values of x where the series (of functions) do not converge, and then demonstrates that the limit of the series has measure zero.

But, convergence almost everywhere has a subset built-in with measure zero. Why is it not possible to simply show that where the function does not converge, the measure is zero (since this is the definition of a.e. convergence)?

Thank you. (Sorry there is nothing explicate written here, I will look up the formatting to write things properly tomorrow).

But on say $\mathbf{R}$ you have the classic example of $\mathbf{1}_{[n,\infty)}$. This converges pointwise to zero. So definitely it conveges a.e. But the set where the values are more than epsilon from the limit is always a large set.