Necessary condition for a function to be odd

Question

Let $f:\mathbb R\to\mathbb R$ such that it holds $$\int_{-a}^af(x)dx=0\qquad \forall a\in\mathbb R^+.$$

Is it true then that the function $f$ is odd, i.e. $f(x)=-f(-x)?$

Answer 1

First, let us define the following function: $$F(a)=\int_{-a}^af(x)dx\qquad \forall a\in\mathbb R^+.$$ If $f$ is continuous $F$ is well defined and we can take the derivative, because of the fundamental theorem of calculus:

$$F'(a)=f(a)+f(-a)$$

But, by hypothesis we have $F\equiv 0$, so $F'\equiv 0$ and then $f(a)+f(-a)=0~\forall~a\in\mathbb{R}^+$, so $f$ is odd.

If $f$ is not continuous @TitoEliatron gave you a counterexample, so the result is not true.

In fact $f$ is 'almost odd' (is odd in all $\mathbb{R}$ but a set of measure $0$). To correct this and have the result for all $\mathbb{R}$ you need $f$ to be continuous.