Jtdylktuy

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)}{nsin(y)} qquad if,,yin {rm I!R}setminus {kpi | kin {rm I!N} } \ phi_{n}(x,y)=n^2 ,,,,,,qquad if,,yin {kpi | kin {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 ${ kpi | kin {rm I!N}}$ 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 $Y={kpi : kinmathbb N}$ then $|Y|=0$ so by Fubini theorem
$$
int_{Omega} |phi_n(x, y)|dxdy = frac 1nint_{Omegasetminus Y}leftlvertfrac{cos x}{sin y}rightrvert dxdy=frac 1nint^pi_0lvertcos xrvert dxint_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=
$$

Now
$$
int_{mathbb Rsetminus Y}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{(i+1)pi}_{ipi}leftlvertfrac{1}{sin y}rightrvert dy=sum^{+infty}_{i=-infty}int^{pi}_{0}leftlvertfrac{1}{sin y}rightrvert dy=+infty
$$

and doesn't converge