2
$\begingroup$

I am reading Rudin's "Real and Complex Analysis", and on pg 15, Rudin writes:

Let $f$ be an extended-real function on a set X.

If $f = g - h, g \geq 0$, and $h \geq 0$, then $f^+ \leq g$ and $f^- \leq h$, where $f^+$ and $f^-$ denote the postive and negative parts of f.

Here, shouldn't Rudin require that both $g(x)$ and $h(x)$ cannot be $+\infty $ at the same time because Rudin has not yet defined $\infty -\infty $?

$\endgroup$
1
$\begingroup$

This is a bit of a subtle point, in my opinion the exact kind that Rudin is likely to omit.

$f$ is well-defined and fixed before $g$ and $h$. In other words, we start with $f$ and we choose extended real functions $g$ and $h$ so that $f = g-h$ makes sense pointwise everywhere. Therefore it is implicit that $g$ and $h$ are not $+\infty$ at the same time, as otherwise we could not satisfy the constraint $f=g-h$ in a meaningful way.

We could run into the problem you are foreseeing if we were to start with $g$ and $h$ and define $f$ to be $g-h$. Fortunately this is not the case here.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.