Integral value of a function is zero implies the set where the function doesn't take zero values is a null set

Question

Let $f ∈ R[a,b]$. Then show that if $\int_{a}^{b}f(x)dx = 0$, $\{x ∈ [a,b] : f(x) \neq 0\}$ is a null set.

I've tried it in this way... Suppose the set isn't null set. Then $\{x ∈ [a,b] : f(x) \neq 0\}$ is uncountable. Then there will be a subinterval which will consist uncountable no. of points where $f$ doesn't take value 0. After that I'm stuck. Please help.

Thanks in advance !!

Answer 1

This is false. Take $f(x)=x$ on $[-1,1]$ for a counter-example.

If $\int_c^{d} f(x)dx=0$ whenever $[c,d] \subseteq [a,b]$ then the conclusion is true.