Limit infimum of integrals on a triangle Let $T=\{(x,y)\in \mathbb{R}^2: 0\leq \lvert x\rvert \leq y\leq 1\}$ and $\mu$ the restriction of the Lebesgue measure in $\mathbb{R}^2$ to $T$. Suppose that $f\in L^2(T,\mu)$ (and hence also in $L^1(T,\mu)$). Show that \begin{equation}\liminf_{y\to 0^+}\int_{-y}^{y} \lvert f(x,y)\rvert\,\text{d}x=0.\end{equation}
Feels like I've tried everything I can think of. It seems to me that I need to demonstrate a sequence $y_n\searrow 0$ such that
\begin{equation}\lim_{n\to 0} \int_{-y_n}^{y_n} f(x,y_n)\,\text{d}x=0.\end{equation}
I have no idea how to pick such a sequence. I also suspect there is probably an easier way to do this using one of the convergence theorems. Any help would be appreciated.
 A: \begin{align*}
\int_{-y}^{y}|f(x,y)|dx\leq\left(\int_{-y}^{y}1dx\right)^{1/2}\left(\int_{-y}^{y}|f(x,y)|^{2}dx\right)^{1/2}=\sqrt{2y}\cdot\left(\int_{-y}^{y}|f(x,y)|^{2}dx\right)^{1/2},
\end{align*}
so
\begin{align*}
\dfrac{\left(\displaystyle\int_{-y}^{y}|f(x,y)|dx\right)^{2}}{2y}\leq\int_{-y}^{y}|f(x,y)|^{2}dx.
\end{align*}
Assuming that
\begin{align*}
\liminf_{y\rightarrow 0^{+}}\int_{-y}^{y}|f(x,y)|dx>\epsilon_{0}
\end{align*}
for some $\epsilon_{0}>0$, then there exists some $\delta_{0}>0$ such that
\begin{align*}
\int_{-y}^{y}|f(x,y)|dx>\epsilon_{0},~~~~0<y<\delta_{0}.
\end{align*}
We have
\begin{align*}
\dfrac{\epsilon_{0}^{2}}{2y}<\int_{-y}^{y}|f(x,y)|^{2}dx
\end{align*}
and hence
\begin{align*}
\int_{0}^{\delta_{0}}\dfrac{\epsilon_{0}^{2}}{2y}\leq\int_{0}^{\delta_{0}}\int_{-y}^{y}|f(x,y)|^{2}dxdy\leq\int_{0}^{1}\int_{-y}^{y}|f(x,y)|^{2}dxdy=\|f\|_{L^{2}(T,\mu)}^{2}<\infty.
\end{align*}
Obviously,
\begin{align*}
\int_{0}^{\delta_{0}}\dfrac{\epsilon_{0}^{2}}{2y}=\infty,
\end{align*}
a contradiction.
