Let $f,g$ be real valued function on a subset $A$ of $\Bbb R^n$. Show that if $\int_A f$ and $\int_A g$ exist then $\int_A fg$ exists.
I proved that if $f$ and $g$ are integrable on a closed interval $I$ then so is $fg$ so far, how do I complete this?
Suppose that $f$ and $g$ are integrable on $A \subset \mathbb R^n$. By definition (see comments under the question), this means that $$\int_A f = \int_{\mathbb R^n} f \chi_A$$ and $$\int_A g = \int_{\mathbb R^n} g \chi_A$$ both exist. I take this to mean that both exist as proper Riemann integrals, so in particular, $f \chi_A$ and $g \chi_A$ are zero outside some closed bounded intervals $J$ and $K$ respectively. Let $I$ be a closed bounded interval containing $J$ and $K$. Then $$\int_{\mathbb R^n} f \chi_A = \int_{I} f \chi_A$$ and $$\int_{\mathbb R^n} g \chi_A = \int_{I} g \chi_A$$ You have already proved the result for the case where the integral is taken over a closed bounded interval, so we can apply that result to the functions $f \chi_A$ and $g\chi_A$ to conclude that $$\int_I (f \chi_A)(g\chi_A)$$ exists. Now note that $$\int_I (f \chi_A)(g\chi_A) = \int_I fg\chi_A = \int_{\mathbb R^n} fg\chi_A = \int_{A} fg$$