Uniform Convergence for Functions $X \to [0,+\infty]$?

Question

An exercise in Tao's Introduction to Measure Theory asks to show that if $(X,\mathcal{B}, \mu)$ is a finite measure space and $f_n : X \to [0,+\infty]$ is a sequence of unsigned $\mathcal{B}$-measurable functions that converge uniformly to a limit function $f : X \to [0,+\infty]$, then $\int_X f_n\,d\mu$ converges to $\int_X f\,d\mu$.

But wait! $[0,+\infty]$ is not naturally a metric space, so I'm not sure how to interpret uniform convergence here. Should I modify the problem to only consider measurable functions $X \to [0,+\infty)$, or is there a way to recover a meaningful statement that applies to all unsigned measurable functions?

Edit: Added finiteness hypothesis

Answer 1

$\newcommand{\artanh}{\operatorname{artanh}}$$[0,+\infty]$ is homeomorphic to $[0,1]$ and any choice of homeomorphism gives a compatible metric. For example, you could use $\rho(x,y)=|\artanh(x)-\artanh(y)|$ (with $\artanh(\infty):=1$) and define uniform convergence in this sense. There is a more general notion involving uniform spaces.

But for the particular theorem at hand, it doesn't really matter. Excise the set on which $f=\infty$ - call the resulting subspace $Y:=X\setminus f^{-1}\{\infty\}$ - and consider the usual notion of uniform convergence for functions valued in $[0,\infty)$; for large enough $n$, $f_n$ should also take finite values if the sequence can be reasonably said to uniformly converge. You may add that as an extra assumption if you wish. Then the theorem will show $\int_Y f_n\to\int_Yf$.

If $\mu(f^{-1}\{\infty\})=0$ then $\int_Yf=\int_Xf$ and $\int_Yf_n=\int_Xf_n$ so there is no difference. If $\mu(f^{-1}\{\infty\})>0$ then $\int_X f$ is infinite and we must show that $\int_X f_n$ diverges too. To have a notion of uniform convergence here, we could just say (as a reasonable but ad hoc notion):

For all $M>0$ there is $N\in\Bbb N$ such that $f_n(x)>M$ if $n>N$ and $x\in f^{-1}\{\infty\}$.

In that case, it is clear that $\int_X f_n$ diverges (using nonnegativity...) and thus the difference between $[0,\infty)$ and $[0,\infty]$ is immaterial. The real content of the theorem lies in handling the finite case.

Note that this ad hoc definition of uniform convergence on $f^{-1}\{\infty\}$ agrees with the notions imposed by metrics like $\rho(x,y)=|\artanh(x)-\artanh(y)|$. The choice of metric should be irrelevant.