real analysis – Rudin Chap 11, Exc 4: proof check

The exercise: If $f$ is integrable on $E$ and $g$ is bounded and measurable on $E$, then $fg$ is integrable on $E$ also.

My proof: Suppose $f geq 0, 0 leq g leq M$. Note that $fg$ is measurable since $f,g$ are both measurable. Because $fg geq 0$, there exists a monotonically increasing sequence of simple functions ${s_n}$ that converges pointwise to $fg$ on $E$. Hence $|s_n(x)| leq M cdot f(x)$ for all $x in E$. By the Dominated Convergence Theorem, $$lim_{ntoinfty} int_E s_n = int_E fg.$$

Since ${s_n}$ is monotonically increasing, it suffices to show that $sup int_E s_n < infty$. Because $s_n$ is in the set of all simple functions $s$ such that $0 leq s leq Mcdot f$, we have that $$int_E s_n leq Mint_E f < infty.$$

From this follows that $Mint_E f$ is an upper bound for ${int_E s_n }$, so $int_E fg leq M int_E f < infty$. (The other cases follow by similar logic by just dividing $f,g$ into their constituent constant-sign components.)

I checked some other solution manuals, but none of them seemed to use a simple function approximation for $fg$. They all use simple functions to approximate $g$, so I was wondering if there was anything incorrect with my proof? I’d be grateful if someone could point out any errors. Thank you!