If $(f_n)_n\rightarrow f$ almost everywhere and in measure, does $(f_n)_n \rightarrow f$ almost uniformly?
I've been wondering about this question and came with no result. I know that convergence almost everywhere does not imply convergence almost everywhere, but I am not aware of any counterexample if the extra condition of convergence in measure is added.
The other implication is true. If $(f_n)_n \rightarrow f$ almost uniformly, then it is correct that it converges for $f$ both almost everywhere and in measure, which makes me think that the converse (the question) is not true. Thank you!
No, consider $f_n = 1_{A_n}$ where we define,
$$ A_n = [0,1/n] + n = \left[n, n + \frac1n \right]. $$
Evidently $f_n \rightarrow 0$ pointwise and for all $\varepsilon >0$ we have $\mu(\{|f_n| > \varepsilon\}) \leq \mu(\{f_n \neq 0\}) = 1/n \rightarrow 0.$ So $f_n \rightarrow 0$ in measure.
Now let $A \subset \mathbb R$ be measurable with $\mu(A)<\infty.$ Since $\mu\left(\bigcup_{n \geq N} A_n\right) = \infty$ for all $N$ (as the harmonic series diverges), there $m \in \mathbb N$ such that $A_n \setminus A \neq \varnothing$ for all $n \geq m.$ But then for all $n \geq m,$ there exists $x_n \in A_n \setminus A$ such that $f_n(x_n) = 1.$ Therefore $f_n$ does not converge uniformly to $0$ on $\mathbb R \setminus A.$