Measurability of $t \mapsto \int_{\Omega(t)}f(t)g(t)h(t)$ given measurability of $t \mapsto \int_{\Omega(t)}f(t)g(t)$? Suppose I know that, given $f(t), g(t) \in L^2(\Omega(t))$,
$$t \mapsto \int_{\Omega(t)}f(t)g(t)$$
is measurable on $([0,T], Lebesgue) \to (\mathbb{R}, Borel)$. Suppose $h(t) \in L^\infty(\Omega(t))$ is a function which is continuous wrt. $t$. Is this enough to conclude that
$$t \mapsto \int_{\Omega(t)}f(t)g(t)h(t)$$
is measurable in the same sense as above? Let $\Omega(t)$ be a bounded smooth domain.

Let $\Omega(t) = [-1,1]$ and $f(t) = 1$.  Let $A$ be a non-measurable set in $[0,1]$, 
$$g(t)(x) = \cases{1 & if $x \ge 0$ and $t \in A$\cr
               -1 & if $x \le 0$ and $t \in A$\cr
                0 & otherwise\cr}$$
Let $h(t) = 1$ on $[0,1]$ and $0$ on $[-1,0)$.  Then $\int_{\Omega(t)} f(t) g(t) = 0$, but $\int_{\Omega(t)} f(t) g(t) h(t)$ is the indicator function of $A$, which is non-measurable. 
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