Proof of $\forall B \!\in\! R\;\forall C\!\in\! R \left\{\forall A\!\in\! R[(0< A <1) \rightarrow (A \cdot B \le C)] \rightarrow (B \le C) \right \}$

Question

I was reading Bartle's "The Elements of Integration and Lebesgue Measure" when I came across the following inference:

This, for me, can be transated as:

$\forall B \!\in\! R\;\forall C\!\in\! R \left\{\forall A\!\in\! R\;[(0< A <1) \rightarrow (A \cdot B \le C)] \rightarrow (B \le C) \right \}$

My question is: Why is this sentence true?

I know $B$ is the supremum of $A \cdot B$, and I have been told that this fact proves the sentence, but I cannot see how.

Answer 1

Think of this set $$ S = \left\{ \alpha \int \varphi \, d\mu : 0 < \alpha < 1 \right\}. $$ The book shows that $$ \lim \int f_n \, d\mu $$ is an upper bound for $S$. Then, as you said $$ \sup S = \int \varphi \, d\mu. $$ By definition, the supremum will be smaller than all other upper bounds. Hence $$ \int \varphi \, d\mu = \sup S \leq \lim \int f_n \, d\mu. $$