Lebesgue integral equals Riemann Integral on $\mathbb{R}$

Question

I know that if a function $f:[a,b]\rightarrow \mathbb{R}$ is Riemann Integrable then it is Lebesgue Integrable there with respect to the lebesgue measure and the integrals coincide. Does this theorem also apply to improper Riemann integrability on unbounded intervals of the real line e.g. over $\mathbb{R}=(-\infty,\infty)$ itself?

Answer 1

No. remember that $f$ is Lebesgue integrable if and only of $|f|$ is Lebesgue integrable. But this isn't true for Riemann (improper) integrability.

So theere may exists functions that are Riemann (improper) integrable but such that its absolute value is not integrable. The classical example is $\dfrac{\sin x}{x}$ over $(0,+\infty)$.

It can be shown that $\int_0^{+\infty}\dfrac{\sin x}{x}$ converges but $\int_0^{+\infty}\dfrac{|\sin x|}{x}$ diverges. It is known as Dirichlet Integral).

One can see this function as the equivalent to the alternating Harmonic series.