Theorem 4.29 in Baby Rudin: Is this analogue correct?

Question

Here is Theorem 4.29 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Let $f$ be monotonically increasing on $( a, b )$. Then $f( x+ )$ and $f( x- )$ exist at every point $x$ of $( a, b )$. More precisely, $$ \sup_{ a < t < x } f(t) = f( x- ) \leq f(x) \leq f( x+ ) = \inf_{ x < t < b } f(t).$$ Furthermore, if $a < x < y < b$, then $$ f( x+ ) \leq f( y- ).$$

Rudin further states

Analogous results evidently hold for monotonically decreasing functions.

Now is the following a correct statement of the analogous result?

Let $f$ be monotonically decreasing on $(a, b)$. Then $f(x+)$ and $f(x-)$ exist at every point $x$ of $(a,b)$. More precisely, $$ \inf_{a

If not, then what is the correct statement of the analogue?

Answer 1

Yes.

One way to check this is to take $g(x)=-f(x)$. Then if $f(x)$ is monotonically decreasing, then $g(x)$ is monotonically increasing, so we apply the original theorem to $g$ and get:

(1) For every $g(x-)=-f(x-)$ and $g(x+)=-f(x+)$ exist, and specifically $$-f(x-)=g(x-)=\sup_{a\leq t

(2) If $a

Answer 2

Yes, that would be correct; just put $-f$ in the first result and use the fact that $\inf (-f) = - \sup f$ and $\sup(-f) = -\inf f$ over the corresponding intervals (these formulas work here because $f$ is monotonically decreasing, so it is bounded below over $(a,x)$ and bounded above over $(x,b)$ for all $a < x < b$). Or more simply, just sketch such a function and it will be intuitively clear that this is indeed the analogous result.