Suppose $f : [a, b] \to \mathbb{R}$ is a bounded function. Prove if $P$ is a partition of $[a, b]$, then $U(f, P) = −L(−f, P)$ and $U(f) = −L(−f)$.

Question

Suppose $f : [a, b] \to \mathbb{R}$ is a bounded function. Prove if $P$ is a partition of $[a, b]$, then $U(f, P) = −L(−f, P)$ and $U(f) = −L(−f)$.

I know that that $f$ being bounded means that for all $x \in [a,b]$ there exists $M$ such that $|f(x)| \leq M$. But I don't know how to use that fact and that $P$ is a partition of $[a,b]$ then $U(f, P) = −L(−f, P)$.

I automatically think of $U(f, P) - L(f, P) < \varepsilon$ but that is only if a function is integrable...

Any help, please?

Answer 1

Let $$U(P, f) = \sum_{i = 1}^{n}M_i\Delta x_i \quad \text{and} \quad L(P, f) = \sum_{i = 1}^{n}m_i\Delta x_i$$ where $M_i = \sup\{f(x) \mid x_{i - 1} \le x \le x_{i}\}$ and $m_i = \inf\{f(x) \mid x_{i - 1} \le x \le x_{i}\}$. For example, suppose on an interval $[x_{i - 1}, x_i]$ that $\sup(f(x)) = 5.$ Then $\inf(-f(x)) = -5$. Therefore, $$ \sup\{f(x) \mid x_{i - 1} \le x \le x_{i}\} = -\inf\{-f(x) \mid x_{i - 1}\le x \le x_{i}\}. $$ Thus observe that if we consider $-f$, we see that $M_i = -m'_i$ where $m'_i = \inf\{-f \mid x_{i - 1} \le x \le x_i\}$. Hence, we have that $$ U(f, P) = \sum_{i = 1}^{n} M_i\Delta x_i = \sum_{i = 1}^{n} -m_i\Delta x_i = -L(-f, P). $$ You can then apply this result to prove the more general statement.