Can I say that $\int f_n\ d\mu \to \int f\ d\mu$?

Question

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.

Answer 1

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

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

Answer 2

Hint: Use Holder's inequality and the fact that your space is of finite measure.

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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