I am new to Royden's Real Analysis and am wondering what the relationship between Lebesgue and Riemann integrals.

For example, are the two integrals $\int_{\mathbb R} 1_{[0,1]}dm$ and $\int _0^1 1 dx$ equivalent? (m is Lebesgue measure on R). If yes, why?

In general, how does one transform a Riemann integral into a Lebesgue integral (when possible) and vice versa?

  • $\begingroup$ What do you mean by "equivalent"? Do you mean "equal"? $\endgroup$
    – Wojowu
    Jul 10, 2016 at 13:33

2 Answers 2


Yes, they are equivalent. Every Riemann integrable function in the standard sence (not in the improper sense) is Lebesgue integrable. This is a standard result in measure theory.

Let $f: [a,b] \to \mathbb R$ a Riemann integrable function. Then you have

$$ \int_a^b f(x) dx = \int_{\mathbb R} f \cdot 1_{[a,b]} dm .$$

This is possible since every Riemann integrable function is Lebesgue integrable and the interval $[a,b]$ is Borel measurable and hence $1_{[a,b]}$ Lebesgue integrable.

Further there are functions that are Lebesgue integrable but not Riemann integrable (an example is the Dirichlet function). But there are also functions that are improper Riemann integrable but not Lebesgue integrable (an example is $\frac{\sin x}{x}$ on $(0, \infty)$, since it is not absolutely improper Riemann integrable).


Yes, the two integrals are equivalent.

The general theorem is: if the Riemann integral exists, then so does the Lebesgue integral, and they are equal.

The relationship is, given $f$ measurable: $$\int_{a}^{b}{f(x)\,dx} = \int_{\mathbb{R}}{1_{[a,b]}(x)\,f(x)\,dL^1 (x)}$$

The vice versa is not guaranteed; you may take as an example $$x \mapsto 1_{[0,1]\cap \mathbb{Q}}(x)\,,$$ which has a finite Lebesgue integral whose value is 0 (because it has nonzero values in a numerable-sized set) but has no Riemann finite integral.

NB At the end of the day, you compute the integral in the same way. With the Lebesgue integral and measure theory though, you have access to several more results.

  • $\begingroup$ I really don't think it's purposeful if you write first a short answer with not much information in it and then copy another existing answer under the same post. $\endgroup$
    – Yaddle
    Jul 10, 2016 at 14:01
  • $\begingroup$ Hey, you are right, I'm sorry. I'm still new to the answer format of this site though, as you can see I did a lot of editing because my answer lacked the requirements of this site. $\endgroup$
    – Eugenio
    Jul 10, 2016 at 14:06

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