Show that $\lim_{n \to \infty} \int_{0}^{1}|f_n(x)| \, dx= 0$ Let $\{f_n\}$ be a sequence of Lebesgue integrable functions and $g: [0,\infty)\to [0,\infty)$ be an increasing and continuous function such that $\displaystyle\lim_{x\to\infty}g(x) = \infty$.
We also have that
(i) $\int_0^1|f_n(x)|g(|f_n(x)|)\,dx < 100$
(ii) $ f_n \to 0$ almost everywhere
We must show that
$$ \lim_{n\to\infty}\int_0^1 |f_n(x)| \, dx = 0$$  
What I have:
Let $\epsilon > 0$,
$g$ is continuous and increasing, then
$$ \exists M > 0, g(|f_n(x)|) > M \mbox{ and } \frac{100}{M} < \frac{\epsilon}{2}$$
Now consider the following partition of $[0,1]$
$$E_1 = \{x \in [0,1]: |f_n(x)| \leq \frac{1}{M}\}$$
$$E_2 = \{x \in [0,1]: |f_n(x)| > \frac{1}{M}\}$$
Then 
$$\int_0^1 |f_n(x)|\,dx = \int_{E_1}|f_n(x)| + \int_{E_2}|f_n(x)| \, dx$$
Thus,
\begin{align}
& \left|\int_0^1|f_n(x)|\,dx\right| < \frac{1}{M}m(E_1) + \int_0^1 |f_n(x)| \frac{g(|f_n(x)|)}{g(|f_n(x)|)} \, dx \\[10pt]
< {} & \frac{1}{M} + \frac{1}{M} \int_0^1 |f_n(x)|g(|f_n(x)|) \, dx < \frac{100}{M} + \frac{100}{M} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.
\end{align}
I want to know if I got the proof right. Thanks.

$\forall \varepsilon>0$
$\exists M>0,\ s.t.\ 100/g(M)<\varepsilon/3$.
Due to Egoroff. $\exists E(measurable)\subset[0, 1]\ \&\ m([0, 1]-E)<\varepsilon/(3M)$. And $f_n\xrightarrow{u.}0$ on E.
Find an enough large N such that $\forall n>N$, $|f_n|<\varepsilon/3$ on E.
Then we have
\begin{eqnarray}
\int_0^1|f_n|dx&=&\int_E|f_n|dx+\int_{[0, 1]-E}|f_n|dx\\
&\leq&\varepsilon/3+\int_{([0, 1]-E)\cap\{|f_n|\leq M\}}|f_n|dx+\int_{([0, 1]-E)\cap\{|f_n|> M\}}|f_n|dx\\
&\leq&\varepsilon/3+\int_{[0, 1]-E}Mdx+\int_{[0, 1]\cap\{|f_n|>M\}}|f_n|dx\\
&\leq&2\varepsilon/3+\int_{[0, 1]\cap\{|f_n|>M\}}|f_n|\frac{g(|f_n|)}{g(|f_n|)}dx\\
&\leq&2\varepsilon/3+\frac{1}{g(M)}\int_0^1|f_n|g(|f_n|)dx\\
&\leq&\varepsilon
\end{eqnarray}
