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There are many kinds of metrics that can induce the topology of convergences in measure. Two most common metrics are

  1. Here is the first one.

$ d(f,g) := \inf_{\delta > 0} \big(\mu(|f-g|>\delta) + \delta\big) $

  1. Here is the second one, which is the most common.

$ d(f,g) = \int \frac{|f-g|}{1+ |f-g|}d\mu $

I have three questions:

Q1 I am trying to prove the Triangle inequality of the first metric. I have done so far:

$ \mu\left\{ x\in X:\left|f\left(x\right)-g\left(x\right)\right|>\delta\right\} +\delta \le \mu\left\{ x\in X:\left|f\left(x\right)-h\left(x\right)\right|>\frac{\delta}{2}\right\} +\frac{\delta}{2}+\mu\left\{ x\in X:\left|h\left(x\right)-g\left(x\right)\right|>\frac{\delta}{2}\right\} +\frac{\delta}{2} $

But I have no idea what to do next.

Q2 What is the difference between the two metrics? Both metrics can induce the topology of convergence in measure. As far as I am concerned, the second metric is used only in probability contexts or in the case the measure is finite. I wonder whether this statement is right. If it is true, what's wrong with the second metric when the measure is not finite?

Q3 Given a measure space $\left(X,\mathscr{F},\mu\right)$ and let $L^{0}\left(X,\mathscr{F},\mu\right)$ be the vector space of all real-valued measurable functions on $\left(X,\mathscr{F},\mu\right)$. If both metrics can be defined on $L^{0}\left(X,\mathscr{F},\mu\right)$, are the topologies of these two spaces the same?

Can anyone help me out? Thanks in advance.

Leoalan.Huang
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1 Answers1

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Q1: let $A=d(f,h)$ and $B=d(h,g)$. Let $\epsilon >0$. Then there exist $\delta_1 >0$ and $\delta_2 >0$ such that $\mu(|f-h| >\delta_1)+\delta_1 <A+\epsilon$ and $\mu(|h-g| >\delta_2)+\delta_2 <B+\epsilon$. Note that $|f-g| >\delta_1+\delta_2$ implies that either $|f-h| >\delta_1$ or $|h-g| >\delta_2$. Hence $d(f,g) \leq \mu(|f-g| >\delta_1+\delta_2\leq A+B+2\epsilon$. Since $\epsilon$ is arbitrary we get $d(f,g) \leq d(f,h)+d(h,g)$.

If $\mu$ is not finite it is not true that all bounded measurable functions are integrable. The second definition cannot be used for infinite measures since metrics cannot take the value $\infty$.

For a finite measure both metrics yield the same topology since $f_n \to f$ in one metric iff $f_n \to f$ in the other.

Kavi Rama Murthy
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  • Thanks for your answer. I will fill out the proof. Do you have any idea on the Q1? – Leoalan.Huang Jul 20 '19 at 15:06
  • @Leoalan.Huang I have edited my answer. – Kavi Rama Murthy Jul 20 '19 at 23:23
  • Brilliant! Thanks very much for your answer. – Leoalan.Huang Jul 22 '19 at 03:44
  • For the second definition, couldn't one just use the metric $\min{d, 1}$, even for an infinite measure? The induced topology should be the same. – Danny Feb 15 '21 at 00:43
  • @DannyHansen $\min {d,1}$ gives a bad metric. The distance between any two different constant functions $f$ $g$ would be $1$, so $\frac 1 n$ would not converge to $0$! – Kavi Rama Murthy Feb 15 '21 at 05:07
  • @KaviRamaMurthy Right, but it's only bad because $d$ is bad (since they are topologically equivalent, indeed they have the same Cauchy sequences). What I'm saying is that the issue isn't that $d$ isn't a finite metric, the issue is that it gives the wrong topology when $\mu$ isn't finite. – Danny Feb 15 '21 at 10:25