$f(x)$ is a bounded function such that $\int_a^b(f(x))dx=0$ then does it imply that $f(x)=0$

Question

The function that I have chosen is $f(x)=1, a \le x \le \frac{a+b}{2},and f(x)=-1 , \frac{a+b}{2} < x \le b $.

The function is bounded and with only one point of discontinuity so it is Riemann integrable

Now, if we choose our partition to be $\{a,\frac{a+b}{2},b\}$ then we see that $U(P,f)=0 $ and $L(P,f)=0$.We know by the property of Riemann integration that if we choose any partition $Q$ on $[a, b]$ then $U(Q,f)\ge L(P,f)$ )I can't really conclude after this.

Answer 1

Not necessarily. Take the function $$f(x)=sgn(x)$$ and take $a=-1 \ and \ b=1$. The integral is zero but the function is not itself zero over the range of integration.