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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 8\\ Bryden Cais}
\medskip

\item\begin{enumerate}
    \item Observe that
    \begin{align*}
        \int_0^1 \int_1^{\infty} (e^{-xy}-2e^{-2xy})\,\mathrm{d}x\,\mathrm{d}y &= \int_0^1 -\frac{e^{-xy}}{y}+\frac{e^{-2xy}}{y}\Bigg]_1^{\infty}\,\mathrm{d}y\\
        &= \int_0^1 \frac{e^{-y}}{y}-\frac{e^{-2y}}{y} \,\mathrm{d}y\\
        \intertext{since $e^{-xy}\rightarrow 0$ for any $y>0$ as $x\rightarrow\infty$, and we can exclude the slice $y=0$ since we only change the
        outer integral by a set of measure 0 }
        &=\int_0^1 \frac{e^{-y}}{y}(1-e^{-y}) \,\mathrm{d}y>0\\
        \intertext{since the integrand is strictly positive on $(0,1]$.}
    \end{align*}

    \item On the other hand
    \begin{align*}
         \int_1^{\infty}\int_0^1 (e^{-xy}-2e^{-2xy})\,\mathrm{d}y\,\mathrm{d}x &= \int_1^{\infty} -\frac{e^{-xy}}{x}+\frac{e^{-2xy}}{x}\Bigg]_0^{1}\,\mathrm{d}x\\
        &= -\int_1^{\infty} \frac{e^{-x}}{x}-\frac{e^{-2x}}{x} \,\mathrm{d}x\\
        &=-\int_1^{\infty} \frac{e^{-x}}{x}(1-e^{-x}) \,\mathrm{d}x < 0\\
        \intertext{since the integrand is strictly negative on $[1,\infty)$.}
    \end{align*}

    \item If $|e^{-xy}-2e^{-2xy}|$ were finite (and hence integrable) then $e^{-xy}-2e^{-2xy}$
    would be integrable.  It would then follow by the Fubini Theorem that the order of the integrals is irrelevant, and that
    the values of (a) and (b) would agree.  Since this is evidently not the case, we conclude that
    $$\int_{[1,\infty)\times [0,1]} |e^{-xy}-2e^{-2xy}|\,\mathrm{d}\mu_{\rm Leb}=\infty.$$

\end{enumerate}

\item Let $E,F$ be measurable subsets of $\R$ of positive measure.  Observe that $\chi_{E}(x)\chi_{F}(x-t)=\chi_{E\cap (F+t)}(x)$
is integrable.  It follows from Fubini's Theorem that we have
$$\int_{\R\times\R} \chi_E(x)\chi_{F}(x-t)\,\mathrm{d}\mu_{\rm Leb}=\int_{\R}\int_{\R}\chi_E(x)\chi_F(x-t)\,\mathrm{d}t\,\mathrm{d}x=\mu(E)\mu(F)$$
since for any $x\in \R$ we have $\int_{\R}\chi_F(x-t)\,\mathrm{d}t=\mu(F)$.
Since $F,E$ have positive measure, we have
$$\int_{\R\times\R}\mu_{E\cap (F+t)}(x)\,\mathrm{d}\mu>0.$$
If the intersection $E\cap (F+t)$ had zero measure for all $t\in \R$ then the above integral would be zero.  Thus, for some $t$,
$\mu(E\cap (F+t))>0$.



\end{enumerate}
\end{document}