So I have this as my homework in Measure Theory and I don't know how to prove it. The question is the following: On the domain $\Omega = (0, \pi)\times {\rm I\!R}$ we have a function defined as: \begin{equation} \phi_{n}(x,y) = \frac{\cos(x)}{n\sin(y)} \qquad if\,\,y\in {\rm I\!R}\setminus \{k\pi | k\in {\rm I\!N} \} \\ \phi_{n}(x,y)=n^2 \,\,\,\,\,\,\qquad if\,\,y\in \{k\pi | k\in {\rm I\!N} \}. \end{equation} Now I need to prove that the sequence \begin{equation} \{\int_{\Omega}|\phi_{n}(x,y)|dxdy\} \end{equation}
is convergent almost everywhere. I know that since $\{ k\pi | k\in {\rm I\!N}\}$ is just a set of disjoint points, its Lebesgue measure is $0$, hence, we don't care about them. I was specifically told that I should use Fatou's lemma which we can apply due to the absolute sign. Thus, since we have sequence of positive functions we have \begin{equation} \int_{\Omega}\lim \inf |\phi_{n}| \leq \lim \inf \int_{\Omega}|\phi_{n}|. \end{equation} Also, we have the inverse inequality for the $\lim \sup$ as well i.e. \begin{equation} \lim \sup \int_{\Omega}|\phi_{n}| \leq \int_{\Omega}\lim \sup |\phi_{n}|. \end{equation} Then, if the $\lim \inf$ and $\lim \sup$ coincide, we have that the sequence is actually converges and we get that \begin{equation} \int_{\Omega}\lim \inf |\phi_{n}| = \lim \int_{\Omega} |\phi_{n}| = \int_{\Omega}| \lim \sup |\phi_{n}|. \end{equation} However, I am stuguling with how to use the properties of $\phi_{n}$ to show that $\lim \inf = \lim \sup$. Any help would be appriciated.
Let $Y=\{k\pi : k\in\mathbb N\}$ then $|Y|=0$ so by Fubini theorem $$ \int_{\Omega} |\phi_n(x, y)|dxdy = \frac 1n\int_{\Omega\setminus Y}\left\lvert\frac{\cos x}{\sin y}\right\rvert dxdy=\frac 1n\int^\pi_0\lvert\cos x\rvert dx\int_{\mathbb R\setminus Y}\left\lvert\frac{1}{\sin y}\right\rvert dy= $$
Now $$ \int_{\mathbb R\setminus Y}\left\lvert\frac{1}{\sin y}\right\rvert dy=\sum^{+\infty}_{i=-\infty}\int^{(i+1)\pi}_{i\pi}\left\lvert\frac{1}{\sin y}\right\rvert dy=\sum^{+\infty}_{i=-\infty}\int^{\pi}_{0}\left\lvert\frac{1}{\sin y}\right\rvert dy=+\infty $$
and doesn't converge