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Let $(X,\mathcal F, \mu)$ be a finite measure space. Let $\{f_n\}_{n \geq 1}$ be a sequence of $\mathcal F$-measurable functions converges to a $\mathcal F$-measurable function $f.$ Let $1 \lt p \lt \infty$ and $\int |f_n - f|^p\ d\mu \to 0.$ Can we say that $\int f_n\ d\mu \to \int f\ d\mu\ $?

What I tried is as follows $:$ $$\left \lvert \int f_n\ d\mu - \int f\ d\mu \right \rvert = \left \lvert \int (f_n - f)\ d\mu \right \rvert \leq \int |f_n - f|\ d\mu.$$

Now can I able to write $$\int |f_n - f|\ d\mu \leq M \int |f_n - f|^p\ d\mu\ $$ for some $M \gt 0.$ Any suggestion regarding this will be highly appreciated.

Thanks in advance.

Anacardium
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2 Answers2

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Just use Holder's inequality. If $\frac{1}{p}+\frac{1}{q}=1$ then:

$|\int\limits_X (f_n-f)|\leq(\int\limits_X |f_n-f|^p)^{\frac{1}{p}}(\int\limits_X 1)^{\frac{1}{q}}=(\int\limits_X |f_n-f|^p)^{\frac{1}{p}}(\mu(X))^{\frac{1}{q}}\to 0$

Mark
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  • So unless the measure is finite then we cannot conclude it. Right? Because $\left ( \mu (X) \right )^{\frac {1} {q}}$ serves the purpose of $M$ here. – Anacardium Jan 17 '21 at 15:31
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    @Anacardium Yes, the statement is false in general measure spaces. For example, if we take the Lebesgue measure on the real line and the sequence $f_n=\frac{1}{n} 1_{[0,n]}$ then $f_n\to 0$ in $L^2$. However, $\int f_n=1$ for all $n$, and so $\int f_n$ doesn't converge to $\int 0=0$. – Mark Jan 17 '21 at 15:36
  • Very nice counter-example. Thank you so much for your kind help. I will accept your answer as soon as the system permits me to do so. – Anacardium Jan 17 '21 at 15:40
  • What you have shown is that this conclusion is not even true for $\sigma$-finite measure spaces. That's pretty interesting. – Anacardium Jan 17 '21 at 15:42
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Hint: Use Holder's inequality and the fact that your space is of finite measure.


Notes 1.

Holder's inequality implies that for $ \frac{1}{q}=\frac1p+\frac1r $, you have $$ \|g\|_q\le (\mu(X))^{1/r}\|g\|_p $$

Here you want $q=1$.

Notes 2. In general, see $L^p$ and $L^q$ space inclusion