3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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 $\mathbb R \setminus A.$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.