# If $f_n \to f$ in $L^p$, then prove that this sequence converges to $f$

Let $f_n \in L^p(\Omega)$ be a sequence that converges to $f$ in $L_p(\Omega)$.

If $\Omega_n$ is a subset of $\Omega$ such that $\displaystyle{\lim_{n\rightarrow \infty}}\Omega_n=\Omega$, prove that $||f_n \chi_{\Omega_n}-f||_p\rightarrow0$.

This is what I have done:

$\displaystyle{\int_\Omega |f_n \chi_{A_n} - f|^p d \mu}=\displaystyle{\int_{\Omega\backslash\Omega_n} |f|^p d \mu}+\displaystyle{\int_{\Omega\cap\Omega_n} |f_n - f|^p d \mu}$

Now, since $\mu({\Omega \backslash \Omega_n})\rightarrow 0$, then the first integral goes to zero. The second integral is less than:

$\displaystyle{\int_{\Omega} |f_n - f|^p d \mu}$

Which goes to zero as well. Therefore, $||f_n \chi_{\Omega_n}-f||_p\rightarrow0$.

• $\mu(\Omega_n\setminus \Omega)$ need not go to zero, consider $\Omega=(0,\infty)$ and $\Omega_n=(0,n)$. Instead use the Dominated convergence theorem, with the sequence $|f\chi_{\Omega_n}|^p\leq |f\chi_{\Omega}|^p\in L^1(\Omega)$. – Jose27 Dec 12 '14 at 21:47
• @Jose27 Thank you! If you write your comment as an answer, I'll set it as the correct one. – Imanol Pérez Arribas Dec 13 '14 at 12:48

$\mu(\Omega\setminus \Omega_n)$ need not go to zero, consider $\Omega=(0,\infty)$ and $\Omega_n=(0,n)$. Use instead the Dominated convergence theorem, applied to the sequence $|f\chi_{\Omega_n}|^p\leq |f\chi_\Omega|^p\in L^1(\Omega)$.