Is this sufficient to conclude $L^{1}$ convergence?

Question

Let $f_{n}:[-1,1]\to \Bbb{R}$ be a sequence of Measurable functions such that $|f_{n}|\leq\frac{1}{x^{4}}$ and $\displaystyle\int_{[-1,1]}(f_{n}(x))^{4}x^{2}\,dx\leq 1$ and $f_{n}$ converges in measure to $0$. Then does $\int_{[-1,1]}f_{n}\,dx \to 0$ ? .

What I can directly see is that as we already have convergence in Measure and a finite measure space, we need uniform integrability to conclude $L^{1}$ convergence (which is stronger than what is being asked for) . Now one of the sufficient conditions for uniform integrability is that $f_{n}$ be uniformly bounded in $L^{1+p}$ for some $p>0$. But I am unable to show this. Even if I consider outside the ball $(-\epsilon,\epsilon)$ then still I have to prove uniform integrability in $(-\epsilon,\epsilon)$ which is same as proving uniform integrability in $(-1,1)$ .

Any hints?

Answer 1

Indeed, you can conclude by uniform integrability: observe that $\lvert f_n\rvert\leqslant x^{-4}$ implies that $\{\lvert f_n\rvert>R\}\subset (-R^{-1/4},R^{-1/4})$ and a use of Hölder's inequality with the exponents $4$ and $4/3$ gives $$ \int \lvert f_n\rvert\mathbf{1}_{\{\lvert f_n\rvert>R\}} \leqslant \int_{-R^{-1/4}}^{R^{-1/4}} \lvert f_n\rvert=\int_{-R^{-1/4}}^{R^{-1/4}} \lvert f_n\rvert \lvert x\rvert^{1/2}\lvert x\rvert^{-1/2}\leqslant\left(\int_{-R^{-1/4}}^{R^{-1/4}}\lvert x\rvert^{-\frac 12\frac 43} \right)^{3/4}. $$

Answer 2

For any $p,q$ with $\frac{1}{p} + \frac{1}{q} = 1$, we can apply Hölder's inequality with respect to the measure $x^2 dx$ to get $$\begin{align*} \int_{-1}^1 |f_n(x)|^{4/p}\,dx &= \int_{-1}^1 |f_n(x)|^{4/p} \frac{1}{x^2} x^2\,dx \\ &\le \left(\int_{-1}^1 |f_n(x)|^4 x^2\,dx\right)^{1/p} \left(\int_{-1}^1 \frac{1}{|x|^{2q}} x^2\,dx\right)^{1/q}\\ &\le 1 \cdot \left(\int_{-1}^1 |x|^{2-2q}\,dx\right)^{1/q}.\end{align*}$$ If we can choose $q < 3/2$, then $2-2q > -1$ so the right side is finite, and if we can choose $p < 4$ then we have a uniform bound on the $L^{4/p}$ norms of $f_n$, which as you note implies uniform integrability. A little algebra shows that this is possible by choosing any $3 < p < 4$, with corresponding $\frac{4}{3} < q < \frac{3}{2}$.

Note that the $|f_n(x)| \le \frac{1}{x^4}$ hypothesis was not needed.