# 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.

• The product of three $L^2$ functions is not necessarily integrable. – Robert Israel Jul 12 '15 at 18:12
• Suppose it is in $L^\infty(\Omega(t))$. – C_Al Jul 12 '15 at 18:13
• Can't imagine why this would be so – David C. Ullrich Jul 12 '15 at 18:15
• Ask yourself: Define $\omega(x):=\int_a^x\eta(s)\mathrm{d}s$ for $\eta\in\mathcal{L}$. Then does it follow $\omega\in\mathcal{L}$? – C-Star-W-Star Jul 12 '15 at 19:25
• The point is that the Lebesgue integral is useless for nonmeasurable functions. (Additivity fails!) – C-Star-W-Star Jul 12 '15 at 19:30

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. 0\$