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Physicist here. I am trying to rigor up some naive steps in my computations. Consider the following $$ \lim_{\delta\to0}\int_{\delta}^adx\,\frac{\cos(bx)-1}{x^{3/2}\log\frac{1}{x}} $$ where $1>a>0$ and $b>0$. I know that we can Lebesgue integrate all Riemann integrable functions, but the one above is an improper integral. Can we still see it as a Lebesgue integral? if so, why?

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    The integrand doesn't change sign. So if the improper Riemann integral exists, it is also the Lebesgue one. – 2017-01-17
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    The integrand function is continuous and behaves like $C\frac{\sqrt{x}}{\log x}$ in a right neighbourhood of the origin. $\frac{\sqrt{x}}{\log x}$ is an integrable function in a right neighbourhood of the origin (no matter if we use the Riemann integral or the Lebesgue one) so there is no issue at all. – 2017-01-17
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    @JackD'Aurizio can you give me a general theorem that asseses when an improper Riemann integral is also Lebesgue? – 2017-01-17
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    If a non-negative continuous function is (improperly) Riemann integrable, it is also Lebesgue integrable. The difference between Riemann- and Lebesgue- integrability is very subtle, it can be almost neglected until someone needs the completeness of a function space. For sure, continuous functions cannot be pathologic, and over any compact interval every Riemann-integrable function is also Lebesgue-integrable (the Lebesgue integral is an extension of the Riemann integral). – 2017-01-17
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    @JackD'Aurizio something that does not require non-negativity? Also, why do you bother about the behaviour in the neighborhood of the origin? what is the rigorous reason for your comment? – 2017-01-17
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    The Riemann sums of bounded functions cannot be too troublesome to handle. Tha chapter of unbounded functions deserves more care: $\frac{1}{\sqrt{x}}$ is integrable in a right neighbourhood of the origin, $\frac{1}{x}$ is not. In the original integral the only troublesome point to handle is the origin: we just have to understand how the integrand function behaves in a neighbourhood to prove it is integrable (or not). – 2017-01-17

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