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As far as I know the Lebesgue integral generalises the Riemann integral. One key ingredient to this generalisation is that the Lebesgue integral partitions the range, not the domain as in the Riemann theory, to obtain subsets of the domain. The sum of measures of the subsets is the evaluation of the (definite) integral. An upshot is that there is more freedom with the Lebesgue integral than with the Riemann integral (such as freely interchanging limits and integral sign, for example)

My question is this: If I'm working with a (definite) integral which is Riemann integrable, then might it be helpful to apply the Lebesgue theory? For example, suppose the integral is proving tricky to evaluate. Could considering the integral as a Lebesgue integral possibly make the evaluation easier? I don't suppose this would help much with an indefinite integral since Lebesgue derives the Fundamental Theorem of Calculus from his theory which is "the same as" that of the Riemann theory.

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I got a fun standard one where Lebesgue integral is trivial but Riemann integration is slightly annoying:

$f(x)=1/q$ on rational $x=p/q$ and $0$ otherwise.

It's a bit of work to even show this is Riemann integrable, but the Lebesgue integral is zero as the function is zero almost everywhere.

This doesn't adequately address the question I am sure but it is worth noting. I think the value in Lebesgue integration is exactly the limit theorems you mention, dominated convergence being the most useful.

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There are Riemann integrable functions, which are not Lebesgue integrable:

Does $ \int_0^{\infty}\frac{\sin x}{x}dx $ have an improper Riemann integral or a Lebesgue integral?

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