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There are a lot of types of convergences that imply other convergences, like almost uniform convergence implies convergence in measure, etc..., but this is the one that I cannot seem to find an answer for. Let me throw the definitions out there:

A sequence $\{f_n\}$ of a.e. Real-valued, measurable functions is said to be $\underline{\text{convergent in measure}}$ if there is a measurable function $f$ such that for all $\epsilon >0$ \begin{equation} \lim_{n\rightarrow\infty} \mu[\{x; |f_n(x)-f(x)|\geq \epsilon\}=0\end{equation}

A sequence $\{f_n\}$ of integrable functions $\underline{\text{converges in the mean}}$ to $f$ if for an integrable function $f$ it holds that $\int |f_n-f| d\mu \rightarrow 0$ as $n\rightarrow \infty$.

My intuition says that convergence in mean does NOT imply convergence in measure, just because the first seems too weak to imply the second, but honestly I have no way of knowing if thats true or how to prove it. Thanks for any help!

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    $\begingroup$ I believe Chebyshev's inequality shows that convergence in mean gives convergence in measure. However, the typewriter sequence shows that convergence in measure does not imply convergence in mean. $\endgroup$
    – user71352
    Commented Dec 31, 2016 at 20:01
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    $\begingroup$ Supplementing user71352's comment: The Markov/Chebyshev inequality in this non-probability context is the observation that, for all $\epsilon>0$, we have $|f_n(x)-f(x)| \geq \epsilon 1\{|f_n(x)-f(x)|\geq \epsilon\}$ for all $x \in \mathbb{R}$, and so $\int |f_n(x)-f(x)| d\mu \geq \int \epsilon 1\{|f_n(x)-f(x)|\geq \epsilon\} d\mu$. $\endgroup$
    – Michael
    Commented Dec 31, 2016 at 20:15
  • $\begingroup$ @user71352. What is the "typewriter sequence"? $\endgroup$
    – DanielWainfleet
    Commented Jan 1, 2017 at 8:35
  • $\begingroup$ @user254665 The typewriter sequence is the sequence of functions $f_{n}:[0,1]\to\mathbb{R}$ defined by $f_{1}(x)=\chi_{[0,1]}(x)$, $f_{2}(x)=\chi_{[0,\frac{1}{2}]}(x)$, $f_{3}(x)=\chi_{[\frac{1}{2},1]}(x)$, $f_{4}(x)=\chi_{[0,\frac{1}{3}]}(x)$, $f_{5}(x)=\chi_{[\frac{1}{3},\frac{2}{3}]}(x)$, $\ldots$ or some variant of this idea. $\endgroup$
    – user71352
    Commented Jan 2, 2017 at 1:43

1 Answer 1

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Suppose $f_n$ does not converge to $f$ in measure. Then for some $\epsilon>0,$ $\mu(\{|f_{n}-f|\ge \epsilon\})$ does not converge to $0.$ This implies there exists $\delta > 0$ and a subsequence $f_{n_k}$ such that

$$ \mu(\{|f_{n_k}-f|\ge \epsilon\})\ge \delta \, \text { for all }k.$$

Thus

$$\int_X |f_{n_k} - f|\,d\mu \ge \int_{|f_{n_k}-f|\ge \epsilon}|f_{n_k} - f|\,d\mu \ge \epsilon \cdot \delta $$

for all $k.$ Therefore $f_{n_k}$ does not converge to $f$ in mean, contradiction.

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