Consider the space $L^1(\mathbb{R} \times (0,1))$ with the Lebesgue measure. Let $f,g \in L^1(\mathbb{R} \times (0,1))$ and assume that $$ \lim_{h \to 0} \frac{f(x+hy,y)-f(x,y)}{h} = g(x,y) $$ in $L^1(\mathbb{R} \times (0,1))$ norm.
I want to prove that $f(\cdot,y)$ is weakly differentiable for almost every $y \in (0,1)$ and $g(x,y) = yf_x'(x,y)$ for almost every $(x,y) \in \mathbb{R} \times (0,1)$.
I argue as follows. We have \begin{align*} \int_a^b \frac{f(x+hy,y) - f(x,y)}{h} dx &= \frac 1h \int_{a+hy}^{b+hy} f(x,y) dx - \frac 1h \int_a^b f(x,y) dx\\ &= \frac 1h \int_b^{b+hy} f(x,y) dx - \frac 1h \int_a^{a+ty} f(x,y) dx \end{align*} for almost every $y$ (say for $y \in Y \subset (0,1)$ with $Y$ having full measure). Hence $$ \lim_{h \to 0} \int_a^b \frac{f(x+hy,y) - f(x,y)}{h} dx = yf(b,y) - yf(a,y) $$ for almost every $a,b \in \mathbb{R}$ (say for $a,b \in X(y) \subset \mathbb{R}$ with $X(y)$ having full measure in $\mathbb{R}$ for every $y \in Y$).
On the other hand, by assumption, $$ \lim_{h \to 0} \int_{Y'} \int_a^b \frac{f(x+hy,y)-f(x,y)}{h} dx dy = \int_{Y'} \int_a^b g(x,y) dx dy $$ for any measurable $Y' \subset Y$.
Now it is enough to show that $$ yf(b,y) - yf(a,y) = \int_a^b g(x,y) dx $$ but it is not obvious to me. How can I prove that?