# Convergence in measure implies uniform convergence on a set of finite measure

Question is :

Does there exist a sequence $$\left(f_k\right)$$ of Lebesgue measurable functions such that $$f_k$$ converges to $$0$$ in measure in $$\mathbb{R}$$ but no subsequence converges uniformly on any subset of positive measure?

See that $$\left(f_k\right)$$ converges to $$0$$ in measure implies there exists a subsequence $$\left(f_{n_k}\right)$$ that converges point wise almost every where to $$0$$.

Then by Egoroff's theorem we can say that on a set of finite measure, $$\left(f_{n_k}\right)$$ converges uniformly almost everywhere to $$0$$.

I am not sure if we can say it converges uniformly and not just almost everywhere.

• 1. By Egoroff's theorem we can say that on a set of finite measure, if $(f_{n_k})$ that converges point wise almost every where to $0$, then $(f_{n_k})$ converges almost uniformly to $0$. Please, note that almost uniform convergence is not the same as uniform convergence almost everywhere. – Ramiro May 7 '16 at 6:08
• 2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly. As a consequence, there is a set of of finite measure where $(f_{n_k})$ converges uniformly to $0$. – Ramiro May 7 '16 at 6:25
• @Ramiro : I understand 1st point but not second point – user311526 May 7 '16 at 7:34
• 2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly to $0$. It means, for all $\varepsilon>0$, there is a set measurable set $A$ such that $m(A)<\varepsilon$ and $(f_{n_k})$ converges o $0$ uniformly on $\mathbb{R}-A$. So the answer to your question is NO, there is not a sequence as you described in your question. (See, for instance, Bartle, Elements of Integration, theorem 7.11 or Halmos, Measure Theory, section 22, theorem D). – Ramiro May 8 '16 at 22:54