# What do we mean when we say that a function $f$ takes the value $\infty$?

What do we mean when we say that a function $f$ takes the value $\infty$?

In measure theory it is common to let mappings take values in the extended real number system.

But still it doesn't make sense to say that $f(x)= \infty$ for any specific $x$. So what is meant with $f : X \mapsto [-\infty, \infty ]$; or the preimage of the set whwer $f =\infty$.

I can see two possibilities, for instance we may say $1/x =\infty$ at $x=0$. Or we could say $f= \infty$ on some set $E$ if $f$ is unbounded here.

• I'm not sure what your confusion is. Do you have a problem with the constant function $f(x)=\infty$? – Eric Stucky Nov 5 '14 at 16:42
• No, of course not. So we never say that unbouded functions, like $f(x) =1/x$ on $(0,1)$ equals $\infty$? – Alexander Nov 5 '14 at 16:44
• when do we use functions such as $f(x)= \infty$? – Alexander Nov 5 '14 at 16:44
• Usually $+ \infty$ is not a number. Thus, when we say that $\lim 1/x$ is $\infty$ when $x=0$, we mean that for every $N$ we can find a $\delta$ such that if $|x - 0| < \delta$, then $1/x > N$. – Mauro ALLEGRANZA Nov 5 '14 at 16:48
• Yes, this I know. Why do we need to include $\infty$ as a number in the range of measurable functions? – Alexander Nov 5 '14 at 16:50

In measure theory, we just allow $\infty$ as a value, because it is convenient to do so.

For example, let $f : \Bbb{R} \to \Bbb{R}$ be a (bounded) function with $\int |f|\, dx < \infty$.

Then, we can just define (without having to worry about convergence questions)

$$g : \Bbb{R} \to [0,\infty], x \mapsto \sum_{n \in \Bbb{Z}} |f(x+n)|.$$

By usual Theorems like monotone convergence, we then see that

$$\int_0^1 g(x) \,dx = \sum_{n} \int_0^1 |f(x+n)| \, dx = \sum_n \int_n^{n+1} |f(y)|\, dy = \int_\Bbb{R} |f|\, dx <\infty.$$

Now, we can conclude a posteriori that $g(x) < \infty$ almost everywhere (and thus the sum converges absolutely almost everywhere).

If we had not allowed functions to possibly take the value $\infty$, we would have had to artificially restrict our assumptions on $f$, or to change the summation in $g$ to a finite summation, ...

Hence, the argument was tremendously simplified by allowing $g$ to take the value $\infty$.

In this case, this had nothing to do with something like

$$\frac{1}{x} \to \infty \text{ for } x \downarrow 0.$$

In these cases, the concept of "taking the value $\infty$" is also not too useful, because if you want to integrate such a function, you can just as well define

$$f:x\mapsto\begin{cases} \frac{1}{x}, & x\neq0,\\ \sqrt{\pi}, & x=0, \end{cases}$$

because the singleton $\{0\}$ is a null-set and thus irrelevant for integration purposes. (We could also have used $\infty$ or $0$ instead of $\sqrt{\pi}$, but as I said this is not the most important application of "taking the value $\infty$").

• This is a great answer indeed! – Alexander Nov 5 '14 at 17:37