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 $?


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.

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