Prove that bounded $p$-norms and convergence a.e implies convergence in $L^1$ Suppose $\mu\left(X\right) < \infty$, $f_n \rightarrow f$ a.e and $p > 1$ is such that for some constant $C>0$, we have 
$$
\|f_n\|_p \leq C,\ \ \text{for each} \ n
$$
Prove that $f_n \rightarrow f$ in $L^1$.
My attempt: I am trying to prove that $f_n \rightarrow f$ in $L^p$. Then the result is obvious from Holder's inequality. 
I am unable to use the given condition that $p$-norms are bounded. I need to apply Dominated convergence theorem, but I'm unable to see it.
 A: By Hölder's inequality and $\sup_n\|f_n\|_p\leqslant C$ we have
$$\|f_n\|_1 = \int |f_n\cdot 1| \leqslant \left(\int |f_n|^p\ \mathsf d\mu\right)^{\frac1p} \left(\int \mathsf d\mu\right)^{1-\frac1p}=\|f_n\|_p  \mu(X)^{1-\frac1p}\leqslant C\mu(X)^{1-\frac1p}<\infty, $$
so that $\sup_n\|f_n\|_1\leqslant C\mu(X)^{1-\frac1p}$, which implies that $\|f\|_1\leqslant C\mu(X)^{1-\frac1p}$ and hence $f\in L^1(\mu)$.
Let $\varepsilon>0$. Since $\mu(X)<\infty$, by Egoroff's theorem, there exists a measurable set $E\subset X$ with $$\mu(E)<\frac\varepsilon{2(C\mu(X)^{1-\frac1p}+\|f\|_1)}$$ and $f_n\to f$ uniformly on $E^c$. Then as $f_n\to f$ a.e., $$\lim_{n\to\infty} \int_{E^c} |f_n-f|\ \mathsf d\mu =0. $$ 
Choose $N$ so that $n\geqslant N$ implies $\int_{E^c} |f_n-f|\ \mathsf d\mu<\frac\varepsilon2$. Then for $n\geqslant N$, 
\begin{align}
\|f_n-f\|_1 &= \int |f_n-f|\ \mathsf d\mu\\
&=\int_{E^c} |f_n-f|\ \mathsf d\mu + \int_{E} |f_n-f|\ \mathsf d\mu\\
&\leqslant \frac\varepsilon2 + \mu(E)\|f_n-f\|_1\\
&< \frac\varepsilon2 + \frac\varepsilon{2(C\mu(X)^{1-\frac1p}+\|f\|_1)}(\|f_n\|_1+\|f\|_1)\\
&\leqslant \frac\varepsilon2+\frac\varepsilon2=\varepsilon,
\end{align}
so that $$f_n\stackrel{L^1}\longrightarrow f. $$
