Full Question: Suppose that $\mu$ is a finite measure. Show that if $\lim_{n \to \infty} f_n = 0$ in $\mu$ - measure and $\limsup_{n \to \infty} \int \vert f_n \vert^2 d\mu <1$, then $\lim_{n \to \infty} f_n =0$ in $L^r (\mu)$ for every $r \in [1,2)$.
This is a question from a prelim paper. I have tried to prove it by contradiction but with no success.
I started off by assuming that there was a subsequence $n_k$ and $\epsilon > 0$ such that $ \int |f_{n_k}| d\mu > \epsilon$.
I let $B_n = \{x \vert f_n(x) > \frac{\epsilon}{\mu(X)}\}$. We know $\mu(B_n) \to 0$ as $n\to \infty$.
I want $\int_{B_{n_k}} \vert f_{n_k} \vert > \beta$ for some $\beta$, as I have $\vert\vert f_{n_k} \chi_{B_{n_k}}\vert\vert_1 \leq \vert\vert f_{n_k}\vert\vert_2 \vert\vert \chi_{B_{n_k}} \vert\vert_2 \leq 1 \times \mu (B_{n_k}) \to 0 $ as $n\to \infty$.
Any suggestions would be greatly appreciated.
Actually, there is no need to use contradiction. Define $B_n$ as $\{x,f_n(x)\gt\varepsilon\}$ (we assume $f_n$ non-negative for simple). Then using Hoelder's inequality, $$\int_X f_n^r\mathrm d\mu\leqslant \varepsilon^r+\left(\int_X f_n^{r\cdot\frac 2r}\mathrm d\mu\right)^{\frac r2}\cdot \mu(B_n)^{1-\frac r2}\leqslant \varepsilon^r+\left(\int_X f_n^2\right)^{r/2}\mu(B_n)^{1-\frac r2},$$ hence $$\limsup_{n\to\infty}\int_X f_n^r\mathrm d\mu\leqslant \varepsilon^r.$$