$f = 0$ outside a set of measure zero implies $\int_a^b f \, dx= 0$

Question

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.

Answer 1

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.

Answer 2

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.