$ \mu (A_y)=0$ for every $ y $ implies $\mu (\bigcup_{1 Assume $ f_n:R\to R$ is continuous
$\{f_n\} $ a sequence of the continuous functions, $n \in N $
And that for a fixed $ y $, $\lim_{n\to \infty} f_n (xy)=f(x,y)$ for almost every $ x $
$ A_y=\{x:\lim_{n\to \infty} f_n (xy) \neq f(x,y)\} $
$ \mu (A_y)=0$,
$\mu $ , Lebesgue outer measure
It doesn't look obvious to me $$\mu (\bigcup_{1 <y <2} A_y)=0$$
One obvious application is that $ f (x, y) $ is measurable over $ X\times Y $
Where $x \in X \subset R$, $y \in Y \subset R$
 A: This is not an answer but a proof of another result that would have been implied by the answer:
Given $f:R\to R$ is measurable
$x \in X \subset R$
$y \in Y \subset R$
Prove $f(xy)$ is measurable over $X\times Y$
Proof :
notation: $3B$ is the set with same center as $B$ but with 3 times the radius. $b\cdot B$ implies each element of $B$ is dilated by a factor $b$, $\mu^*$ 1-dimensional Lebesgue outer measure over $R$,and $\mu$ 2-dimensional Lebesgue outer measure over $R^2$
There is a sequence of continuous functions {$f_n$} such that for every $y$ , $\lim_{n \to \infty}f_n(xy)=f(xy)$ for almost every $x$
Now we just need to show $\lim_{n \to \infty}f_n(xy)=f(xy)$ for almost every point in $X\times Y$
it suffice to prove $\lim_{n \to \infty}f_n(xy)=f(xy)$ for almost every point in $(c,d)\times (a,b)$, where $c,a >\delta>0$
Let $A_y=\{x :\lim_{n \to \infty}f_n(xy)\neq f(xy)\}$,$\mu^*(A_y)=0$ $\frac {1}{2}>\epsilon >0,n,i,j \in N$, $j>i$
Let $I_{y,n}=(a_{y,n},b_{y,n})$ be an open interval such that $a_{y,n}>\delta>0$, $$A_y \bigcap (c, d)\subseteq \bigcup_{n=1}^{n=\infty} I_{y,n}$$, and $$\mu^*(\bigcup_n I_{y,n})< \epsilon$$
$$B_{y,i,j}=\bigcup_{n=i}^{n=j} I_{y,n}$$
$$C_{y,j}=\bigcup_{n=j}^{n=\infty} I_{y,n},\lim_{j \to \infty}\mu^*(C_{y,j})=0$$
$$L_{y,j}=\min_{n \le j} {\frac{b_{y,n}}{a_{y,n}}}$$
$$B_{b,i,j}=\frac{1}{b}\cdot B_{1,i,j},    B_{a,i,j}=\frac{b}{a}\cdot B_{b,i,j}$$
$$\mu^*(B_{a,i,j})=\frac{b}{a}\mu^*( B_{b,i,j})$$
it follows that $$L_{ay,j}=L_{y,j}$$
Step 1: take $i=1$ and $j=m$ such that $\mu^*(B_{y_1,1,m}) < \epsilon$ and $\mu^*(C_{y_1,m}) < \epsilon^2$
Step 2: make the partition $y_1>b>y_2>y_3>...y_{k-1}>a>y_k$ such that $\frac{y_n}{y_{n+1}}=L_{y_1,m}$ ,ie $y_{n}=(L_{y_1,m})^{-n+1}y_1$ , creating $k-1$ segments.
observe  that : $$\bigcup_{y_{n+1} <y<y_n} B_{y,1,m}\times(y_{n+1},y_n)\subseteq 3B_{y_n,1,m}\times(y_{n+1},y_n)$$
let the total number of segments be  $N_{y_1,1,m}=k-1$
$$N_{y_1,1,m}=\frac{ln(y_1)-ln(y_k)}{ln(L_{y_1,m})}$$
$$\bigcup_{y_k <y<y_1} B_{y,1,m}\times(a,b)\subseteq \bigcup_{n=1}^{n=k-1} 3B_{y_n,1,m}\times(y_{n+1},y_n)$$
Now repeat step 1 and 2 now with $i=m,j=z$ such that $\mu^*(B_{y_1,m,z}) < \epsilon^2$ and $\mu^*(C_{y_1,z}) < \epsilon^4$
and get : $$\bigcup_{y_k <y<y_1} B_{y,m,z}\times(y_{n+1},y_n)\subseteq \bigcup_{n=1}^{n=k-1} 3B_{y_n,m,z}\times(y_{n+1},y_n)$$
observe  that : $$ \lim_{z \to \infty} L_{y_1,z}=1$$ ,
$$\mu(\bigcup_{n=1}^{n=k-1} 3B_{y_n,m,z}\times(y_{n+1},y_n)) \le 3\epsilon\sum_{n=1}^{n=k-1}(y_n-y_{n+1}) L_{y_1,z}^{n-1}=3\epsilon y_1\sum_{n=1}^{n=k-1}L_{y_1,z}(L_{y_1,z}-1)=3\epsilon y_1(k-1)L_{y_1,z}(L_{y_1,z}-1)=3\frac{\epsilon y_1L_{y_1,z}(L_{y_1,z}-1)(ln(y_1)-ln(y_k))}{ln(L_{y_1,z})}=3\epsilon K_{y_1,z}$$
$L_{y_1,z}=1+p$, and since $L_{y_1,z}>1$,$p>0$
using taylor's expansion of $ln(1+p)$, it can easily be seen $\frac{L_{y_1,z}-1}{ln(L_{y_1,z})}$ , is bounded by some non negative constant even as $p \to 0$
so there is a constant $M$ such that $|y_k - y_1| < M$,  $K_{y_1,z} < M$ for all $z$
$$\bigcup_{m<z}\bigcup_{y_k <y<y_1} B_{y,m,z}\times(a,b)\subseteq \bigcup_{m<z}\bigcup_{n=1}^{n=k-1} 3B_{y_n,m,z}\times(y_{n+1},y_n)$$
$$D=\{(c, d) \times (a,b) :\lim_{n \to \infty}f_n(xy)\neq f(xy)\}$$
$$ D \subseteq \bigcup_{m<z}\bigcup_{y_k <y<y_1} B_{y,m,z}\times(a,b)$$
$$\mu(D)\le \sum_{i=1}^{i=\infty}3M\epsilon^i \le 6M\epsilon$$
