Let $\{f_n\}$ be a sequence of measurable functions on $R$ with Lesbegue measure and $f$ be a measurable function.

I have to show that $\{f_n\}$ converges to $f$ in measure if and only if any subsequence of $\{f_n\}$ has a subsequence that converges to $f$ almost everywhere. It was not difficult to show the "only if" part. But I am stuck at the "if" part. Could anyone help me with this?


Suppose $\{f_n\}$ converges to $f$ in measure. Let $\{f_{n_k}\}$ be a subsequence of $\{f_n\}$. Choose $n_{k_1} < n_{k_2} < n_{k_3} < \cdots$ such that $m(|f_{n_{k_j}} - f| > 2^{-j}) < 2^{-j}$ for all $j \in \Bbb N$. Let $A_j = (|f_{n_{k_j}} - f| > 2^{-j})$ and $A = \limsup A_j$. Since $\sum_j m(A_j) < \infty$,

$$m(A) = \lim_{j\to \infty} m\left(\bigcup_{k\ge j} A_k\right) \le \lim_{j\to \infty} \sum_{k = j}^\infty m(A_k) = 0.$$

Given $x\notin A$, there exists a positive integer $N$ such that $x\notin A_j$ for all $j\ge N$, i.e., $|f_{n_{k_j}}(x) - f(x)| \le 2^{-j}$ for all $j \ge N$. Thus $f_{n_{k_j}}(x) \to f(x)$. As $x$ was arbitrary, $f_{n_{k_j}} \to f$ a.e..

Now suppose $\{f_n\}$ does not converge to $f$ in measure. Decompose $\Bbb R$ as the countable union of open intervals $I_k$. Then there exists an index $\ell$, positive numbers $\epsilon$ and $\alpha$, and a sequence of indices $n_1 < n_2 < n_3 < \cdots$ such that $m(\{x\in I_\ell :|f_{n_k}(x) - f(x)| > \epsilon\}) \ge \alpha$ for all $k \in \Bbb N$. Let $\{f_{n_{k_j}}\}$ be any subsequence of $\{f_{n_k}\}$. If $X$ is the set of points $x\in I_\ell$ such that $\lim\limits_{j\to \infty} f_{n_{k_j}}(x) \neq f(x)$, then $X = \cup_{m\ge 1} (\limsup\limits_{j\to \infty} X^j_m)$, where $X_m^j = \{x\in I_\ell :|f_{n_{k_j}} - f| > \frac{1}{m}\}$. Let $N$ be a positive integer such that $\frac{1}{N} < \epsilon$. Then

$$ m(X) \ge m(\limsup\limits_{j\to \infty} X_j^N) \ge \limsup_{j\to \infty}\, m(X_j^N) \ge \alpha.$$

It follows that $\{f_{n_{k_j}}\}$ does not converge to $f$ pointwise almost everywhere.

  • $\begingroup$ This proves the "only if" part. Wasn't OP stuck on the "if" part? $\endgroup$ – grand_chat Apr 9 '15 at 15:43
  • $\begingroup$ Could you explain more about the last part of your answer? Why is the measure of limsup X_(N,j) larger than α? $\endgroup$ – Keith Apr 27 '15 at 13:17
  • $\begingroup$ But, in oder for the inequality to hold, the union of X_(N, j) must be of finite measure. $\endgroup$ – Keith Apr 27 '15 at 14:09
  • $\begingroup$ Thank you for your explanantion. I was stuck at that part. $\endgroup$ – Keith Apr 27 '15 at 14:10
  • $\begingroup$ Wait! does the assumption that the measure of X is zero implies that the union of X_(N, j) is of finite measure? I think that doesn't have to be true. $\endgroup$ – Keith Apr 27 '15 at 14:32

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.