Let $f \colon [a,b] \to \mathbb R$ bounded, such that $f(x) = 0$ for every $x \in [a,b]$ except in a set $J$ of measure zero. When we say that a set $J$ has measure zero, if given any $\varepsilon > 0$ there exist a countable collection of open intervals $( a_n ,b_n)$ such that $J \subset \bigcup\limits_{n \in \Bbb N} (a_n ,b_n)$ and $\sum \limits_{n \in {\Bbb N}} (b_n - a_n) < \varepsilon$. Prove that in the Riemann sense (I don't know any other sense of integrals) the integral exist and $$ \int_a^b f (x) \, dx = 0. $$

The existence is easy, but how can I prove the equality? Help me with this please. Don't use Lebesgue integrals, because I can't use it in this exercise. It's from a real analysis course. Thanks!

I know that this result it's more general, Instead of putting $f(x) = 0$, I can put any Riemann integrable function, but the general case, comes off as trivial corollary of this. So why try to prove this.

  • $\begingroup$ For general knowledge, Lebesgue integrals are considered to be a very (very) main theme in real analysis. $\endgroup$
    – Shai Deshe
    Dec 1, 2011 at 16:25
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    $\begingroup$ If $f(x) = \begin{cases} 1 & \text{if }x\text{ is rational}, \\ 0 & \text{if }x\text{ is irrational}, \end{cases}$ then that seems to satisfy all of the hypotheses even though $f$ is not Riemann-integrable. $\endgroup$ Dec 1, 2011 at 16:26

2 Answers 2


I would've thought the existence would be harder.

Assuming you already proved the existence, let $\varepsilon>0$ and let $M$ be the bound on $f$.

You can cover $J$ with a set of intervals whose accumulated length is less than $\varepsilon$.

If you assume that all the intervals are pairwise disjoint, you get that the integral is bound from above by $\varepsilon M$, and relaxing that demand can only reduce the value of the integral.

This asserts that the value of the integral can be bound from above by $\varepsilon M$ for any $\varepsilon >0$, as needed.

  • $\begingroup$ @Michael Hardy Maybe, I must add the hypothesis of being integrable $\endgroup$
    – August
    Dec 1, 2011 at 16:52

For the reason Michael Hardy indicated in a comment, the statement is incorrect. If you assume that the function is Riemann integrable, then Shai Deshe's answer applies.

But you're correct that if both $f$ and $g$ are Riemann integrable and $f=g$ almost everywhere, then $\int_a^b f(x)\,dx=\int_a^b g(x)\,dx$. This could be shown directly using reasoning similar to that in Shai Deshe's answer.

You mention that the initial problem is a trivial corollary of the more general case (at least I think that's what you meant). But it turns out that the general case is also a trivial corollary of the special case! If $f(x)=g(x)$ almost everywhere and both $f$ and $g$ are Riemann integrable, then $h(x)=f(x)-g(x)$ is Riemann integrable, and $h(x)=0$ almost everywhere. So $\int_a^b h(x)\,dx=0$, and since $\int_a^b h(x)\,dx=\int_a^b f(x)\,dx-\int_a^b g(x)\,dx$, the general result follows.

  • $\begingroup$ P.S: "almost everywhere" is another way to say "outside of a set of measure zero". I forgot when writing my answer that that terminology was not used in the question. $\endgroup$ Dec 17, 2011 at 6:36

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