Let $f_n$ be a sequence of measurable functions which converge in measure to a function $f$. My first question is, is $f$ itself necessarily measurable?
Now suppose that $|f_n| \leq |g|$ for some integrable function $g$. How do we argue that $|f-f_n|$ is integrable? If $f_n$ converged pointwise to $f$, then we would also have $|f| \leq |g|$, so that $|f- f_n| \leq 2 |g|$, but we can't do that here. The textbook I'm using is Royden, by the way.
Yes the limiting function is measurable. To prove this, show that convergence in measure implies convergence almost surely along a subsequence. Then use the fact (or reprove easily) that almost sure convergence implies measurability. As a hint, recall that you would need to show $\{x: \ f(x)\leq t\}$ is measurable for each $t$. If you focus on the set $A$ where $f_n(x)$ converges to $x$, then then you can write $\{x: \ f(x)\leq t\}$ as $\cap_n \{x: f_n(x)\leq t\}$ plus some measure 0 set.
To se that $|f-f_n|$ is integrable, use triangle inequality: $|f-f_n|\leq |f|+|f_n|\leq |f|+|g|\leq 2|g|$, where the r.h.s. is integrable. You are using the fact that if $|f|\leq |g|$, where $|g|$ is integrable, then so is $f(x)$ (assuming both are measurable of course).