General condition that Riemann and Lebesgue integrals are the same

I'd like to know that when Riemann integral and Lebesgue integral are the same in general. I know that a bounded Riemann integrable function on a closed interval is Lebesgue integrable and two integrals are the same.

However, I want to know as general as possible condition for two integrals to be the same. For example, in terms of improper integrals, unbounded functions and R^n space.

Any answer would be helpful. Thank you.

Solution 1:

Although this doesn't fully satisfy all your requests, I think it will be useful to state a theorem which appears in some similar form in Folland's Real Analysis (I believe 2.28, though I might be wrong):

Theorem: Suppose that $f:[a,b] \to \Bbb{R}$ is bounded. Then:

1) If $f$ if Riemann integrable, then $f$ is Lebesgue measurable and the Riemann integral $\int_a^b f(x) \, dx$ equals the Lebesgue integral $\int_{[a,b]} f \, d\mu$ (where $\mu$ is Lebesgue measure).

2) Further $f$ is Riemann integrable if and only if the set of discontinuities of $f$ is a $\mu$-null set.

End Theorem

Let me make another point about a difference in Riemann and Lebesgue integration from a geometric standpoint. If you notice, the Lebesgue integral $\int_{[a,b]} f \, d\mu$ has no "orientation" with respect to the interval $[a,b]$, whereas the Riemann integral $\int_a^b f(x)\,dx$ we interpret as "the integral of $f$ from $a$ to $b$" and, in fact, we have $\int_a^b = - \int_b^a$ for Riemann integrals. To further investigate this point, I'd suggest looking into differential geometry; in this context $dx$ would play the role of the "Riemann volume form" on $\Bbb{R}$ which depends on this orientation, whereas $d\mu$ is an honest measure and doesn't require orientation.

Solution 2:

The Riemann integrable functions are a subset of the upper functions $\mathscr L^+(I)$, those functions which are a increasing limit a.e. on $I$ of a sequence of step functions, hence usually a proper subset of the Lebesgue integrable functions $\mathscr L(I)=\mathscr L^+(I)-\mathscr L^+(I)$. That is, whenever $f$ is Riemann integrable over $I$ it is also Lebesgue integrable, and the integrals coincide, but there are functions which are Lebesgue integrable but not Riemann integrable.

You must be careful, however, since the Riemann integral is only defined for blocks in $\Bbb R^n$; i.e. those bounded compact subsets that are products of intervals $[a,b]$, and we may stretch the definition up to Jordan measurable sets, however one can define the Lebesgue integral of a function over any Lebesgue measurable subset of your space. Again, a Jordan measurable set is Lebesgue measurable, but there are Lebesgue measurable sets that are not Jordan measurable.

As a counterpart, the positive improperly Riemann integrable functions are Lebesgue integrable, but there are improperly Riemann integrable functions that are not Lebesgue integrable. The usual (counter)example is $x\mapsto x^{-1}\sin x$ over $I=\Bbb R$.

Solution 3:

Your statement about "a bounded Riemann integrable function" is correct, but you don't need the "bounded" part: an unbounded function can't be Riemann integrable. It can have an improper Riemann integral, but that's a different matter.

Lebesgue's criterion for Riemann integrability: A function on a bounded interval $I$ is Riemann integrable if and only if it is