Understanding the case $\alpha < 0$ in thm. (5) page 75 in Royden "Fourth Edition".

Question

The theorem and part of its proof are given below:

My question is:

In case of the proof in the case $\alpha < 0$: How is the proof related to our previous knowledge that $\inf A = -\sup (-A)$, could anyone explain this for me, please?

Answer 1

Using the facts $\inf A = - \sup(-A)$ and $\sup \lambda A = \lambda \sup A$ where $\lambda A := \{\lambda a: a \in A\}$ and $-A := (-1)A$, we have

$$\inf_{\varphi \geqslant \alpha f}\int_E \varphi = \inf_{[\varphi/\alpha] \leqslant f}\alpha \int_E [\varphi/\alpha] = - \sup_{[\varphi/\alpha] \leqslant f}(-\alpha)\int_E [\varphi/\alpha] = -(-\alpha)\sup_{[\varphi/\alpha] \leqslant f}\int_E [\varphi/\alpha] \\= \alpha \sup_{[\varphi/\alpha] \leqslant f}\int_E [\varphi/\alpha]$$

Here we use $ -\alpha > 0$ and $ \varphi \geqslant \alpha f \implies \varphi/\alpha \leqslant f$ when $\alpha < 0$.