Equality on Riemann and Lebesgue integral (basic question)

Question

Let $f,g:\mathbb{R}\to\mathbb{R_+}$ two real functions so that $\int f(u)g(u)du$ exists. Does $$\int f(u)g(u)du=\int fgd\mu \quad ?$$ where $\mu$ is the Lebesgue measure (if not, what conditions should be imposed to make it true?). Secondly, if I suppose $g$ with compact support, say $supp \ g=[-a,a]$, then $$\int f(u)g(u)du=\int_{-a}^a f(u)g(u)du=\int_{[-a,a]} fgd\mu \quad ?$$

Can you help me to clarify these points?

This question came from a book I'm reading. The author presented a Riemann integral and then stated results about $\mu$-almost everywhere related to that Riemann integral. So, it is clear that he treated the Riemann integral as Lebesgue integral. I just want to make the equivalence clear. It is important to mention that, in the case, $g$ is symmetric, bounded, nonnegative, has compact support, and $f$ is a polynomial.

Thanks in advance!

Answer 1

Any Riemann integrable function on $[a,b]$ is Lebesgue integrable and the two integrals are the same . However we cannot say the same about improper Riemann integrals. A well known example is $\frac {\sin x} x$ on $[0,\infty)$.

Answer 2

Addition to Kabo's answer. In the example $\int\sin(x)/x\;dx$ change variables to get the example $$ \int_0^1\frac{\sin(1/x)}{x} dx $$ where the improper Riemann integral exists, but the Lebesgue integral does not. The problem is that the OP does not say his functions are continuous.

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As noted in a comment from Celine Harumi: if a nonnegative function is improperly Riemann integrable, then it is Lebesgue integrable. So this is what the OP needs.