Show f is Riemann integrable if L(-f,[a,b])=-L(f,[a,b])

Question

This question is from Sheldon Axler's real analysis book exercise 1b. Suppose f is a bounded function on [a, b]. Show that f is Riemann integrable on [a,b] iff L(-f,[a,b])=-L(f,[a,b]).

This is how I Done.

Suppose f is Riemann integrable, then $\int_a^b -f \,dx$ = L(-f, [a, b]) = sup$\Sigma$inf(-f)$(x_(i+1)-x_i)$. Noticing that sup(-f) = -inf(f) and inf(-f) = -sup(f). We have sup$\Sigma$inf(-f)$(x_(i+1)-x_i)$ = -inf$\Sigma$sup(f)$(x_(i+1)-x_i)$ = -U(f, [a,b]) = - $\int_a^b f \,dx$.

Thus, we have - $\int_a^b f \,dx$ = $\int_a^b -f \,dx$ = L(-f, [a,b]) = -L(f, [a, b]) which shows one direction.

I think the other direction is False. Consider the function f(x) = 0 if x is rational and 1 if x is irrational. This function is not Riemann integrable while L(-f, [a, b]) = -L(f, [a,b]) = 0. Not sure if the counterexample is valid. If the statement is true, I'm not sure how to prove it.

Answer 1

Observe that $L(-f,[a,b])=-U(f,[a,b]).$ Therefore the equality $L(-f,[a,b])=-L(f,[a,b])$ implies $U(f,[a,b])=L(f,[a,b]),$ hence the function $f$ is Riemann integrable.

Answer 2

For the function You gave, the $L(f,[a,b])=0$, but $L(-f,[a,b])=-1$,

@Ryszard gave the proof for general case... showing that the upper sum and Lower sum are equal and hence, f is differentiable.