Anywhere I integrate $f_n$, the integral approaches $f$. Is $\lim_n f_n = f$ a.e.? Something tells me this is obvious...
I have a bunch of functions: $f,f_n:\mathbb{R}^2\rightarrow \mathbb{R}$, all integrable. Also, $f$ is continuous. 
I also have a family of sets, $\mathcal{G}$ where $\sigma(\mathcal{G})=\mathcal{B}(\mathbb{R}^2).$ (the Borel sigma-algebra of $\mathbb{R}^2$, perhaps sans some sets of measure 0 wrt Lebesgue)
Now the important part: we have that $\int_{S} f_n d\lambda \rightarrow \int_{S} f d\lambda$ for any $S\in \mathcal{G}$.
Does $\lim_n f_n = f$ a.e.?

Here is my thought: 
Think of this set: $S:=\{\lim_n f_n \neq f\}.$ What would happen if:
$$\lambda(S)>\varepsilon > 0?$$
Then $\lim_n \int_S |f_n-f| = \lim_n \int_S (f_n-f)^+ +\int_S (f_n-f)^- > 0$. But both of these parts must go to 0 by assumption. So we have a contradiction.
 A: What you are trying to show is false. To see this, consider
$$
f_n (x,y) = \chi_{[0,1]^2} \cdot e^{2\pi i \langle (n,0), (x,y)\rangle}.
$$
For an arbitrary measurable set $S \subset \Bbb{R}^2$, we have
$$
\theta_n := \int_S f_n \, d(x,y) = \langle e^{2\pi i \langle (n,0), (x,y)\rangle}, \chi_{[0,1]^2 \cap S}\rangle_{L^2([0,1]^2)}.
$$
But $(e_{n,k})_{n,k}=(e^{2\pi i \langle (n,k), (x,y)\rangle})_{n,k \in \Bbb{Z}}$ is an orthonormal basis of $L^2([0,1]^2)$, so that we have
$$
\infty > \Vert \chi_{[0,1]^2 \cap S} \Vert_{L^2([0,1])}^2 = \sum_{n,k \in \Bbb{Z}} |\langle \chi_{e_{n,k}, [0,1]^2 \cap S}\rangle|^2.
$$
Together, we easily conclude $\int_S f_n \, d(x,y) = \theta_n \to 0 = \int_S f \, d(x,y)$ as $n \to \infty$,  for $f \equiv 0$.
But $|f_n| \equiv 1 $ on $[0,1]^2$, so that $f_n \to f$ almost everywhere does not hold.
I am aware that I am using complex valued functions instead of real valued ones. But one can easily adapt the above argument (use ${\rm Re}f_n$ instead of $f_n$) to the real-valued case.
EDIT: Instead of using the orthonormal basis property, one can also use the Riemann Lebesgue Lemma to conclude $\theta_n \to 0$.
