I am trying to show that if $f_n$ converges to $f$ in $L^p(X,\mu)$ then $f_n\to f$ in $L^p$ in measure, where $1\le p \le \infty$.

Here is my attempt for $p>1$ - Let $\varepsilon>0$ and define $A_{n,\varepsilon}=\lbrace x: \vert f_n(x)-f(x) \vert \ge \varepsilon\rbrace$. I want to show $\mu (A_{n,\varepsilon})\to 0$. $\Vert f_n-f \Vert_p=(\int _X \vert f_n-f\vert ^p)^{1/p}\ge \int _{A_{n,\varepsilon}}(\vert f_n-f\vert ^p)^{1/p}\ge \varepsilon \mu (A_{n,\varepsilon})^{1/p}$ so that $\mu (A_{n,\varepsilon})\le (\frac{\Vert f_n-f\Vert }{\varepsilon})^{1/p}$ and the RHS tends to $0$ as $f_n\to f$ in the $L^p$ norm.

How can I deal with the case $p=\infty$?

  • 1
    $\begingroup$ Perhaps I'm missing something, but shouldn't the $L^{\infty}$ case be the simplest? Since $||f_n-f||_{\infty}<\epsilon \implies \mu(|f_n-f|\geq \epsilon) = 0$. $\endgroup$ – Shalop May 14 '15 at 5:17

The case $p = \infty$ is the simplest since $\|f_n-f\|_{\infty} < \epsilon$ means that $|f_n-f|$ is less than $\epsilon$ almost everywhere, and therefore $\mu\big(|f_n-f|\ge\epsilon\big) = 0$.

If $1\le p < \infty$, you can use Tchebychev's inequality: $$ \mu\big(|f_n-f|\ge\epsilon\big) = \mu\big(|f_n-f|^p\ge\epsilon^p\big) \le \frac{1}{\epsilon^p}\int|f_n-f|^p\,d\mu = \frac{1}{\epsilon^p}\|f_n-f\|_p^p, $$ where the last expression can be made arbitrarily small by taking $n$ sufficiently large.


If the underlying measure has finite variation then convergence in L-infinity implies convergence in Lp and the result follows. However, this is not true for non-finite measures. The constant functions x->1/n defined on the real line converge in L-infinity but not in measure.


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