Show that $\underline{\int}_a^b(-f(x)) \, dx=-\overline{\int}_a^bf(x) \, dx$

Question

If f be bounded function on $[a,b]$, prove that $$\underline{\int}_a^b(-f(x)) \, dx=-\overline{\int}_a^bf(x) \, dx$$

Please help me to solve this problem.

Answer 1

For any subinterval $I_j = [x_{j-1},x_j]$ of a partition $P$ we have $\inf_{x \in I_j}[-f(x)] = -\sup_{x \in I_j}[f(x)]$

Forming lower and upper sums we have

$$L(P,-f) = \sum_{j=1}^n\inf_{x \in I_j}[-f(x)](x_j - x_{j-1}) = -\sum_{j=1}^n\sup_{x \in I_j}[f(x)](x_j - x_{j-1}) = - U(P,f) \\ \implies \underline{\int}_a^b (-f(x)) \, dx = \sup_P L(P,-f) = \sup_P [-U(P,f)] = - \inf_P U(P,f) = -\overline{\int}_a^b f(x) \, dx $$