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!

  • $\begingroup$ Here is a partial answer which doesn't even need convergence in measure. Only that the set where the series is defined has finite measure: proofwiki.org/wiki/Egorov%27s_Theorem $\endgroup$ – Maik Pickl Aug 14 '17 at 12:31

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 $\mathbb R \setminus A.$

  • $\begingroup$ Exactly, I was just about to type this up. Have my upvote. :) $\endgroup$ – Maik Pickl Aug 14 '17 at 12:46
  • $\begingroup$ @ctoi dont you mean $\mu(\{|f_n| > \varepsilon\})$? $\endgroup$ – Marios Gretsas Aug 14 '17 at 12:50
  • $\begingroup$ @MariosGretsas Indeed I do. Thank you for the correction, I've edited my answer accordingly. $\endgroup$ – ktoi Aug 14 '17 at 12:58
  • $\begingroup$ @MaikPickl yes you are right..i really forgot about the Egorov's theorem ..so to give a valid counterexample i need to define an appropriate sequence in an unbounded set for example? $\endgroup$ – Marios Gretsas Aug 14 '17 at 13:02
  • $\begingroup$ @MariosGretsas Yes, exactly as in this answer. :) Which of course is inspired by your, now deleted, answer. :) $\endgroup$ – Maik Pickl Aug 14 '17 at 13:05

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