## Do you have to pay chap fees for withdrawal of binary trading profit?

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## 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!