Show not Lebesgue integrable using step functions I have shown that conditionally $\int_0^\infty \int_0^\infty f(x,y) dx dy = \int_0^\infty \int_0^\infty f(x,y) dx dy = \frac{\pi}{2}$ where $f(x,y) = e^{-xy} \sin(x).$ This part is relatively easy because 
$$\int_0^\infty e^{-xy} \sin(x) dy = \frac{\sin(x)}{x} $$
Now I want to show that $f \not\in L^1(\mathbb{R}^{+2})$.  I know we can show $f$ is not Lebesgue integrable by finding a sequence of simple / step functions $\phi_n \leq |f|$ where $\int_{\mathbb{R}^{+2}} \phi_n \to \infty$, but I need some help here.

It is readily apparent that $f$ is not Lebesgue integrable since $F(x) = \int_0^\infty e^{-xy} |\sin x| \, dy = \frac{|\sin x|}{x}$ is not integrable over $[0,\infty)$.  If $f$ were integrable, the iterated integral must be finite by Tonelli's theorem.
Alternatively, using your suggested approach, take $A_{jk} = \left[\frac{\pi}{4} + j\pi, \frac{3\pi}{4} + j\pi\right] \times \left(\frac{1}{k+1},\frac{1}{k}\right]$ and define the sequence of step functions
$$\phi_{mn}(x,y)= 2^{-1/2}\sum_{k=1}^m \sum_{j=0}^n e^{-\pi\left(\frac{3}{4} + j\pi\right)\frac{1}{k}}\chi_{A_{jk}}(x,y)$$
Since $|\sin x| \geqslant 2^{-1/2}$ for $x \in \left[\frac{\pi}{4} + j\pi, \frac{3\pi}{4} + j\pi\right] $, we have for $(x,y) \in A_{jk}$,
$$e^{-xy} |\sin x| \geqslant 2^{-1/2}e^{-\left(\frac{3\pi}{4} + j\pi\right)\frac{1}{k}}$$
Thus,
$$\begin{align}\int_0^\infty \int_0^\infty e^{-xy} |\sin x| \, dx \, dy  &\geqslant \lim_{m \to \infty}\lim_{n \to \infty} \int_0^\infty \int_0^\infty \phi_{mn}(x,y) \, dx \, dy \\&= 2^{-1/2}\sum_{k=1}^\infty \sum_{j=0}^\infty e^{-\left(\frac{3\pi}{4} + j\pi\right)\frac{1}{k}}\left(\frac{1}{k} - \frac{1}{k+1} \right) \\ &= 2^{-1/2}\sum_{k=1}^\infty \sum_{j=0}^\infty e^{-\frac{\pi j }{k}}\frac{e^{-\frac{3\pi}{4k}}}{k(k+1)}\\ &= 2^{-1/2}\sum_{k=1}^\infty  \frac{1}{1-e^{-\frac{\pi  }{k}}}\frac{e^{-\frac{3\pi}{4k}}}{k(k+1)} \\ &=2^{-1/2}\sum_{k=1}^\infty  \frac{e^{\frac{\pi}{4k}}}{k(k+1)\left(e^{\frac{\pi}{k}}-1 \right)} \\ &= + \infty \end{align}$$
The series on the RHS diverges because the summand is $\sim \frac{1}{\pi (k+1)} $ as $k \to \infty$.
