Suppose we have four Lebesgue measurable functions: $$ f_1: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \qquad f_2: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \\ g_1: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \qquad g_2: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} $$ such that $f_1, f_2$ are almost-everywhere equal by the Lebesgue measure on $\mathbb{R} \times \mathbb{R}$ and $g_1, g_2$ are almost-everywhere equal by the Lebesgue measure on $\mathbb{R} \times \mathbb{R}$.
Are the functions $$ h_1(x, z) = \int_{y \in \mathbb{R}} f_1(x, y)g_1(y, z)\ dy \\ h_2(x, z) = \int_{y \in \mathbb{R}} f_2(x, y)g_2(y, z)\ dy $$ also almost-everywhere equal by the Lebesgue measure on $\mathbb{R} \times \mathbb{R}$?
The answer is yes, at least if you assume everything to be integrable. $f_1,f_2$ can also be viewed as functions from $\mathbb{R}^3$ to $\mathbb{R}$ that happen to not depend on $z$; notice that they are equal almost everywhere w.r.t. the Lebesgue measure on $\mathbb{R}^3$. Similarly for $g_1,g_2$. Then $f_1g_1$ and $f_2g_2$ are functions on $\mathbb{R}^3$ that agree almost everywhere, and we can use Fubini.