I am studying real analysis and I am being introduced to functions that take values on the extended real line, I have a fundamental doubt about this so I'll give an example to illustrate my confusion:

Consider the proposition "If $f$ is measurable then there exists a sequence $(\phi_n)_{n \in \mathbb N}$ of simple functions such that $\phi_n(x) \to f(x)$ for all $x$." So, take the case where there is $x$ with $f(x)=+\infty$, what it means $\phi_n(x) \to f(x)$ when $n \to \infty$?, I suppose the natural definition would be the same as for limits in $\mathbb R$, which is for any $M>0$, there is $n_0:$ for all $n \geq n_0$, $\phi_n(x)>M$. How can I define a metric here?, what would be $d(x,+\infty)$?, which are the open and closed sets?

Thanks in advance


The extended real line is not a metric space. You are correct in noting that $d(0,+\infty)$ must be larger than any $d(0,x)$ for any $x\in\Bbb R$, and therefore cannot be a real number.

However the extended real line is metrizable. This means that we can redefine the metric completely, while preserving the same topology (and therefore the same convergent sequences). To see this simply note that if $\Bbb R$ and $(0,1)$ are homeomorphic, then the extended real line is homeomorphic to $[0,1]$.

But more directly, how do define the open sets? The same way as before, using open intervals, only here you should note that $(x,+\infty]$ and $[-\infty,x)$ are considered an open interval as well, since the endpoints of the entire order are allowed to be included in open intervals.

  • $\begingroup$ By "redefine the metric completely," do you mean redefine the usual Euclidean metric on $\Bbb R$ such that the new metric is a metric on the extended reals resulting in the topology you described? $\endgroup$ – layman Apr 15 '15 at 21:19
  • $\begingroup$ By redefine the metric completely, I mean we can find a different metric, which may or may not agree [on some parts] with the usual Euclidean, absolute value induced metric on $\Bbb R$. $\endgroup$ – Asaf Karagila Apr 15 '15 at 21:20
  • $\begingroup$ Is it possible for this new metric to distort the order, i.e., for three elements $x, y, z$, can we have that $x$ and $y$ are closer than $x$ and $z$ under the usual Euclidean metric, but the opposite is true under the new metric? Probably not if the new metric induces the topology that induces the Euclidean subspace topology on $\Bbb R$. $\endgroup$ – layman Apr 15 '15 at 21:23
  • $\begingroup$ Sure, it's possible. It might stretch a bit the interval $(x,y)$ and then squeeze the rest into being very very small. $\endgroup$ – Asaf Karagila Apr 15 '15 at 21:29
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    $\begingroup$ Sorry, then I didn't understand your original question, and thought you meant that $1$ and $2$ were further than they are in the Euclidean metric, but $2$ and $3$ were closer, or something like that. No, of course you can't change this way, because then $3$ would be in an open ball around $1$ which does not contain $2$, which is impossible if we didn't change the topology. $\endgroup$ – Asaf Karagila Apr 15 '15 at 21:40

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