Let $E⊂\{(x,y)|0≤x≤1, 0≤y≤x\}, E_x =\{y|(x,y)∈E\}, E^y =\{x|(x,y)∈E\}$ and assume that $m(E_x) ≥ x^3$ for any $x ∈ [0, 1].$ Let $E⊂\{(x,y)|0≤x≤1, 0≤y≤x\}, E_x =\{y|(x,y)∈E\},
E^y =\{x|(x,y)∈E\}$ and assume that $m(E_x) ≥ x^3$ for any $x ∈ [0, 1].$
(a) Prove that there exists $y ∈ [0,1]$ such that $m(E^y) ≥ \frac{1}{4}.$
(b) Prove that there exists $y ∈ [0,1]$ such that $m(E^y) ≥ c,$ where $c > \frac{1}{4}$ is a
constant independent of $E.$ Give an explicit value of $c.$
I am having trouble with this real analysis qual problem (past exam).  I was thinking Fubini?  But no luck so far.
 A: First note that the second claim easily implies the second one. We will thus only handle part (b).
Using Fubini's theorem, we calculate
$$
m_{2}\left(E\right)=\int_{0}^{1}m_{1}\left(E_{x}\right)\, dx\geq\int_{0}^{1}x^{3}\, dx=\frac{x^{4}}{4}\bigg|_{0}^{1}=\frac{1}{4},
$$
where $m_{2}$ and $m_{1}$ denote $2$- and $1$-dimensional Lebesgue
measure.
Choose an arbitrary $\alpha\in\left(0,1\right)$. Using Fubini in
the "other direction", and the fact that 
$$
E\subset\left\{ \left(xy\right)\middle|0\leq x\leq1\text{ and }0\leq y\leq x\right\} ,
$$
which implies $E^{y}\subset\left[y,1\right]$, we arrive at
$$
\begin{eqnarray*}
\frac{1}{4} & \leq & m_{2}\left(E\right)=\int_{0}^{1}m_{1}\left(E^{y}\right)\, dy\\
 & = & \int_{\alpha}^{1}m_{1}\left(E^{y}\right)\, dy+\int_{0}^{\alpha}m_{1}\left(E^{y}\right)\, dy\\
 & \leq & \int_{\alpha}^{1}m_{1}\left(\left[y,1\right]\right)\, dy+\alpha\cdot\sup_{y\in\left[0,\alpha\right]}m\left(E^{y}\right)\\
 & \leq & \int_{\alpha}^{1}1-y\, dy+\alpha\cdot\sup_{y\in\left[0,1\right]}m\left(E^{y}\right)\\
 & = & 1-\alpha-\frac{y^{2}}{2}\bigg|_{\alpha}^{1}+\alpha\cdot\sup_{y\in\left[0,1\right]}m\left(E^{y}\right)\\
 & = & \frac{1}{2}-\alpha+\frac{\alpha ^{2}}{2}+\alpha\cdot\sup_{y\in\left[0,1\right]}m\left(E^{y}\right),
\end{eqnarray*}
$$
and thus
$$
\sup_{y\in\left[0,1\right]}m\left(E^{y}\right)\geq-\frac{1}{4\alpha}+1-\frac{\alpha}{2}.
$$
You can now either optimize in $\alpha$ (this will yield $\alpha=\frac{1}{\sqrt{2}}$),
or just take $\alpha=\frac{6}{10}$. For this value of $\alpha$,
the right hand side is
$$
\frac{17}{60}>\frac{15}{60}=\frac{1}{4}.
$$
