Prove if $\{f_n\}$ converges in measure to f and g, then f = g a.e. In my class notes there is a really brief sketch for this proof. Basically by the triangle inequality $\forall \epsilon > 0$ and $\forall k \in \mathbb{N}$:
$m(\{x:|f(x)-g(x)|>\epsilon\}) \le m(\{x:|f(x)-f_k(x)|>\frac{\epsilon}{2}\}) + m(\{x:|f_k(x)-g(x)|>\frac{\epsilon}{2}\})$
which implies $m(\{x:|f(x)-g(x)|>\epsilon\}) = 0$ so f=g a.e.
I was hoping to fill in some of the details, in particular, applying the limits in the definition of convergence in measure. To begin with, by triangle inequality if x is such that $|f(x)-g(x)|> \epsilon$ then $x \in \{x:|f(x)-f_k(x)|>\frac{\epsilon}{2})\cup\{x:|f_k(x)-g(x)|>\frac{\epsilon}{2}\})$ for any $k$. Then by sub-additivity we can get the expression I had in my notes:
$m(\{x:|f(x)-g(x)|>\epsilon\}) \le m(\{x:|f(x)-f_k(x)|>\frac{\epsilon}{2}\}) + m(\{x:|f_k(x)-g(x)|>\frac{\epsilon}{2}\})$
But to conclude the LHS = 0, it seems like I need to take the limit as $n \rightarrow \infty$ to utilize the definition of convergence in measure. Is it valid to just take the limit on both sides? Is any of this right? Thanks for your time!
 A: Yes, it's correct (but given how late it is where I am at the moment, don't trust me completely!). 
The statement $f=g$ a.e. is equivalent to $m(\{x:f(x)\neq g(x)\})=0$; let's denote $N=\{x:f(x)\neq g(x)\}$. (So, we wish to prove that $m(N)=0$.)
You've shown, by taking the limit as $k\to \infty$ in the inequality you obtained, that $m(N_k)=0$ for all natural numbers $k$ where $N_k=\{x:\left|f(x)-g(x)\right|>\frac{1}{k}\}$. 
Of course, $N=\bigcup_{k=1}^{\infty} N_k$ and the countable union of sets of measure zero also has measure zero (by subadditivity of the measure).
I hope this helps! 
A: Yes, it is valid to take the limit as $k\to\infty$ on both sides. 
The left-hand side does not even depend on $k$, so this goes unchanged when taking the limit. The right-hand side is a sequence obtained by adding two convergent sequences together (both of them having limit $0$). So basically, the right-hand side is of the form $a_k+b_k$ with $\lim_k a_k=\lim_k b_k=0$ and hence $\lim_k (a_k+b_k)=\lim_k a_k+\lim_k b_k=0$ by e.g. this.
