Show $\inf (f_n )=- \sup (-f_n)$

Question

In Avner Friendman's Mondern Analysis book, he makes a statement that has stumped me when proving that for $\{f_n\}$ a sequence of measurable functions, $\sup_n f_n$ And $\inf f_n$ Are measurable.

The assertion for $\inf_n f_n $ follows from $\inf (f_n )=- \sup (-f_n)$ .

I really struggle with the concept of inf and sup of a sequence of functions, so I do not see why this statement is true. Why does $\inf (f_n )=- \sup (-f_n)$? Thank you!

Answer 1

Label the $\inf(f_n)=f_I$ and $\sup(-f_n)=f_S$. By definition, $f_I\leq f_n, \forall n$ which implies $-f_I\geq -f_n, \forall n$.

By definition of the sup, $f_S=\inf\{f:f\geq -f_n, \forall n \}$. That is, $f_S$ is the smallest function that is bigger than the sequence $\{-f_n\}_n$. Thus, $-f_I\geq f_S$.

Similarly, $f_S\geq -f_n, \forall n \implies -f_S\leq f_n, \forall n$. By definition of the inf, $f_I=\sup\{f:f\leq f_n, \forall n \}$. That is, $f_I$ is the largest function that is smaller than the sequence $\{f_n\}_n$. Thus, $-f_S\leq f_I$.

Putting the two inequalities together, $-f_S=f_I$.

Answer 2

Hint: Suppose $r\in\mathbb{R}$ is a least upper bound of a set $S\subset\mathbb{R}$ relative to $\mathbb{R}$. Can you show that $-r$ is a greatest upper bound of $-S$? The converse is identical.