Compute limit of integral

Let ${f_{n}}$ be a sequence in $L^{2}(X,\mu)$ such that $||f_{n}||_{2} \rightarrow 0$ as $n \rightarrow \infty$. How to show that:

$\displaystyle \lim_{n \to \infty} \int_{X} |f_n(x)| \log(1+|f_{n}(x)|) d\mu = 0$

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Use the fact that $\log (1+y) \leq y$.

So we get $\displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \leq \displaystyle \int_{X} |f_n(x)| |f_n(x)| d \mu = \displaystyle \int_{X} |f_n(x)|^2 d \mu = ||f_n||_2^2$.

Also, note that $\displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \geq 0$.

Hence, $0 \leq \displaystyle \lim_{n \rightarrow \infty} \displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \leq \lim_{n \rightarrow \infty} ||f_n||_2^2 = 0$

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But in order to carry integration on the LHS we need first to show that the LHS is integrable but it is because it is measurable and it less or equal than a function which is integrable (by assumption), so by a standard result the LHS is integrable, right? –  student Nov 19 '10 at 23:15
@user10: Yes. If you have a measurable function $f$ and another measurable function $g$, then $g(f)$ is also measurable. Hence LHS is measurable and then you can argue as I said. –  user17762 Nov 19 '10 at 23:18
thank you very much. –  student Nov 19 '10 at 23:45