Problem $3.18$, Rudin's RCA (Convergence in Measure) 
Let $\mu$ be a positive measure on $X$. A sequence $\{f_n\}$ of complex measurable functions on $X$ is said to converge in measure to the measurable function $f$ if to every $\epsilon > 0$ there corresponds an $N$ such that
$$\mu(\{x: |f_n(x) - f(x)| > \epsilon) < \epsilon$$
for all $n > N$. Assume $\mu(X) < \infty$ and prove the following statements:

*

*If $f_n(x)\to f(x)$ a.e., then $f_n\to f$ in measure.

*If $f_n \in L^p(\mu)$ and $\|f_n-f\|_p\to 0$, then $f_n\to f$ in measure; here $1\le p\le \infty$.

*If $f_n\to f$ in measure, then $\{f_n\}$ has a subsequence which converges to $f$ a.e.

Investigate the converses of $(a)$ and $(b)$. What happens to $(a)$, $(b)$, and $(c)$ if $\mu(X) = \infty$, for instance, if $\mu$ is the Lebesgue measure on $\mathbb R$?

My (incomplete) work:

*

*I used Egoroff's theorem to complete the proof.


*Fix some $\epsilon > 0$. My approach is similar to the one above: I'm trying to show that $\mu(\{x: |f_n(x) - f(x)| > \epsilon) \xrightarrow{n\to\infty} 0$. (Completed for $1 \le p < \infty$ with help of the hint in the answer. What about $p = \infty$?)


*I see that this part has some discussion here. (Completed).
Follow-up Problem: Are the converses of (a) and (b) true? If not, what are some counterexamples? What happens if $\mu(X) = \infty$?

Comment: In many places, I've seen convergence in measure defined as: $f_n\to f$ in measure if for every $\epsilon > 0$, $$\lim_{n\to\infty} \mu(\{x: |f_n(x) - f(x)| > \epsilon) = 0$$
While this implies the definition used by Rudin, I do not see how the two are equivalent (i.e. why does the other implication hold).
Thanks!
 A: The solution to your comment is the following: we wish to show that the sequence $a_n^\epsilon = \mu \left( x\in X \, : \, |f_n(x) - f(x)| \geq \epsilon \right)$ converges to zero. So let $\delta > 0$, and let's construct an $N = N(\delta )$ such that for all $n \geq N$ we have $|a_n^\epsilon| < \delta$.
First, get $N_1$ from Rudin's definition of convergence in measure, so that for all $n \geq N_1$ we have $a_n^\epsilon < \epsilon$. Next, set $\rho = \min (\epsilon , \delta )$. Get $N_2$ from the definition of convergence in measure with $\rho > 0$ and set $N = \max (N_1, N_2)$. Then, by monotonicity of measure, for all $n \geq N$, $$|a_n^\epsilon| = \mu \left( x\in X \, : \, |f_n(x) - f(x)| \geq \epsilon \right) \leq \mu \left( x\in X \, : \, |f_n(x) - f(x)| \geq \rho \right) < \rho \leq \delta$$
This implies $\mu \left( x\in X \, : \, |f_n(x) - f(x)| \geq \epsilon \right) \to 0$ as $n \to \infty$.
I'll use this equivalent definition to give you some hints for the three questions you asked:

*

*Consider the descending sets $E_N =  \{ x \, : \, \exists n > N \text{ s.t. }|f_n(x) - f(x) | \geq \epsilon \}$. Since $\mu (E_1) < \infty$, continuity from above gives us $$\lim_{N \to \infty} \mu (E_N) = \mu \left( 
 \bigcap_{N=1}^\infty E_N \right)$$ Convince yourself this last measure equals zero since $f_n \to f$ a.e. Finally, what is the relationship between $E_N$ and $\{ x \, : \, |f_n(x) - f(x)| \geq \epsilon \}$ for $n > N$? Also, look up Egoroff's theorem to see that something stronger actually holds).



*Fix $1 \leq p < \infty$, let $E_{n, \epsilon} = \{x \, : \, |f_n(x) - f(x)|^p \geq \epsilon\}$. Consider the inequality $$\int_X |f_n - f|^p \geq \int_{E_{n,\epsilon}} |f_n - f|^p \geq \int_{E_{n,\epsilon}} \epsilon = \epsilon \mu (E_{n,\epsilon})$$ and use this to get convergence in measure.
If $p = \infty$, since $\mu (X) < \infty$ we have $L^\infty (X) \subset L^p (X)$, so the result follows by the above.


*The discussion you linked to contains a complete proof, which makes it a super-hint!
Edit: In response to your follow-up problem, here are a few (partial) converses you can explore:

*

*On a countable atomic measure space, convergence in measure implies convergence almost everywhere. Moreover, on $(\mathbb{N}, \mathcal{P}(\mathbb{N}), \gamma)$, that is counting measure on $\mathbb{N}$, convergence in measure is equivalent to uniform convergence.


*If $(f_n)$ converges in measure to $f$ and $f_n$ is dominated by an integrable majorant $g$, then $f_n \to f$ in $L^1$
Counterexamples when $\mu (X) = \infty$

*

*Consider the indicator $\chi_{[n,n+1]}$ on $\mathbb{R}$ with Lebesgue measure


*For $1 \leq p < \infty$, the result also holds when $\mu (X) = \infty$
