1) Suppose that $A_n$ is measurable with $\mu(A_n) < \infty$ for all $n$ and $1_{A_n}$ converges in measure to $f$, then there exists a measurable set $A$ such that $f = 1_A$
My idea in doing this problem is :
Since $1_{A_n}$ converges in measure to $f$, there exist a subsequence $1_{A_{n_j}}$ converging to $f$ a.e. I think that I should set $A = \cap_{j \in \mathbb{N}} A_{n_j}.$ It is clear that $A$ is measurable. Anyway, I have 2 concerns :
Is $A$ empty ? Should I care if $A$ is empty ? (Empty is a measurable set, but it is strange to write $f = 1_\phi$)
It seem to me that $1_A = f$ a.e. But I find that I am not sure how to prove this rigorously (I have a sense that I might need Dominated convergence thm for this)
2) Change of variable : If $f$ is an integrable function on reals and $a$ is non-zero real number, then $$\int_{\mathbb{R}}f(x+a)\ dx = \int_{\mathbb{R}}f \ dx$$ $$\int_{\mathbb{R}}f(ax) \ dx = |a^{-1}|\int_{\mathbb{R}}f \ dx.$$
I do not sure how to begin. Since $f$ is reals, I should first prove the result for non-negative function. Then it seems that I need to establish this result for an indicator function, then simple function. But it looks complicated. Could anyone give a hint or outline the proof ?
Thank you very much