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I have a background in mathematical statistics and now am trying to self-study topology (from John Kelley's General Topology). I read about convergence classes and how convergence classes generate topologies, and sought to apply the concept to the different notions of convergence in probability theory. Namely, almost sure convergence implies convergence in probability which implies convergence in distribution (or weak convergence). $L_p$ convergence implies convergence in probability for integrable random variables. (There is no relationship between almost sure and $L_p$ convergence in general.) In the end I seemed to reach a contradictory result that makes me question my understanding.

It seems to me that if I convergence classes $\mathscr{C}_1$ and $\mathscr{C}_2$ and $\mathscr{C}_1 \to \mathscr{C}_2$ (meaning convergence in $\mathscr{C}_1$ implies convergence in $\mathscr{C}_2$), then the topology induced by $\mathscr{C}_1$, $\mathscr{T}(\mathscr{C}_1)$ is coarser than the topology induced by $\mathscr{C}_2$, $\mathscr{T}(\mathscr{C}_2)$. That's because a closed set $\bar{A}$ according to $\mathscr{T}(\mathscr{C}_1)$ consists of all points that are limits of nets taking values in $A$, and because all those nets are convergent in $\mathscr{C}_1$, they are also convergent in $\mathscr{C}_2$. Hence, $\bar{A}$ is also a closed set according to $\mathscr{T}(\mathscr{C}_2)$, so every closed set in $\mathscr{T}(\mathscr{C}_1)$ is a closed set in $\mathscr{T}(\mathscr{C}_2)$. Because closed sets and open sets identify each other (one is the complement of the other), every open set in $\mathscr{T}(\mathscr{C}_1)$ is an open set in $\mathscr{T}(\mathscr{C}_2)$, and so $\mathscr{T}(\mathscr{C}_1) \subseteq \mathscr{T}(\mathscr{C}_2)$; $\mathscr{T}(\mathscr{C}_2)$ is finer than $\mathscr{T}(\mathscr{C}_1)$.

With this, a.s. (or $L_p$) convergence has the coarsest topology for random variables, and weak convergence the finest.

Then I show that the topology induced by weak convergence is not in general $T_0$, or a Kolmogorov space, on the space of random variables. Assume the existence of two random variables, $X$ and $Y$, that are not equal to each other on a non-empty set of probability zero but equal otherwise, a possibility generally allowed in probability theory. These two random variables have the same probability distribution (meaning $\mathbb{P}(X\in A) = \mathbb{P}(Y\in A)$ for measurable $A$), and one can then show that $X_n \to X \iff X_n \to Y$ in distribution. Hence, no open set can contain one but not the other, since a closed set with one of these as a member must have the other as a member as both are limit points of anything converging in distribution to one of these. Since every closed set contains both or none of these, the same is true for all open sets. Hence, the induced topology is not $T_0$.

(As an aside, the topology for probability measures is at least $T_2$ on the space of probability measures, and I think that's even true for almost sure and $L_p$ convergence too.)

Because I have shown that the finest topology for random variables is not $T_0$ in general, it should follow that the coarser topologies are not $T_0$ in general either. The finer topology has all open sets appearing in the coarser topologies, so the property found applies for the coarser topologies as well.

This is where I encounter a problem. Almost sure convergence and $L_p$ convergence are both implied by pointwise convergence, or $X_n \to X$ if $X_n(\omega) \to X(\omega)$ for all $\omega \in \Omega$. This convergence seems like it could distinguish the aforementioned $X$ and $Y$; just study a set where $X$ and $Y$ differ, which should exist by hypothesis. But pointwise convergence should induce a coarser topology than even almost sure convergence, and therefore weak convergence.

So I seem to have reached a contradiction. The most likely places where I've made errors are going to be my conclusions about the relationships of the topologies induced by related notions of convergence, as the proof is all my own and I should not trust myself. Less likely is I could also have incorrectly concluded that the topology induced by weak convergence is not $T_0$ when it actually is. Or perhaps pointwise convergence has an incomporable topology (which I also doubt).

cgmil
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  • All topological spaces are convergence spaces, but not all convergence spaces are topological spaces. The characterization of topological convergence spaces is given by Kelley. – Jakobian Jun 23 '25 at 10:07
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    The second paragraph is backwards. If $\mathscr{C}1$-convergence implies $\mathscr{C}_2$-convergence, then the topology generated by $\mathscr{C}_1$-convergence is _finer than the topology generated by $\mathscr{C}2$-convergence. For the closure of $A$ with respect to the latter _contains the closure of $A$ with respect to the former, and it can be strictly larger. There may be nets in $A$ $\mathscr{C}_2$-converging to some points which are not $\mathscr{C}_1$-limits of nets in $A$. – Dermot Craddock Jun 23 '25 at 10:09
  • @DermotCraddock That does not seem to show any relationship between the two topologies other than closed sets in one being subsets of closed sets in the other. But my understanding is that such a relationship does not make one topology finer than the other, since one topology needs to be a subset of the other. In some sense, they're incomparable, if this is all we have. – cgmil Jun 23 '25 at 17:06
  • Well, so far what I've said only implies that if the topologies are comparable, then $\tau_1$ is finer [not necessarily strictly] than $\tau_2$, that's true. But we also have the fact that a $\tau_2$-closed set is $\tau_1$-closed, too. For let $F$ be $\tau_2$-closed, and $y$ a point in the $\tau_1$-closure of $F$. Then there is a net $(x_i)$ that $\mathscr{C}_1$-converges to $y$. But then $(x_i)$ also $\mathscr{C}_2$-converges to $y$, so $y \in F$. – Dermot Craddock Jun 23 '25 at 17:24

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Many notions of convergence in probability theory are not topological. See this answer.

The simplest example is convergence in distribution: take random variables defined on different probability spaces! The boolean valued random variable $X_A$ that it will rain in city $A$ tomorrow and the boolean valued $X_B$ that the minimum temperature in city $B$ will be below freezing have the same Bernoulli distribution if the probability $p$ happens to be the same. An infinite sequence that alternates $X_A$ and $X_B$ converges in distribution to the Bernoulli distribution with parameter $p$ though the two probability spaces are entirely different.

Convergence in distribution has other weird properties. For any symmetric distribution (for example, Gaussian with mean zero, uniform over $[-1,1]$, equal probabilities ($p=1/2$) on the discrete set $\{-a,a\}$), the sequence of random variables alternating between $X$ and $-X$ converges in distribution. It may look like $a$ and $-a$ are being identified, but if the probability is anything other than $1/2$, then they have to be distinguished.

Jayanth R Varma
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  • The answer you cite does not explain why almost sure convergence is not a topological notion of convergence. Can you explain? – cgmil Jun 23 '25 at 14:44
  • The linked answer does not explain why almost-sure convergence is not topological, but an article from the American Mathematical Monthly does: "Convergence Almost Everywhere is Not Topological", E. T. Ordman, AMM 1966. (Link here.) – cgmil Jun 24 '25 at 03:19
  • @Jakobian The simplest example is convergence in distribution: take random variables defined on different probability spaces! The boolean valued random variable $X_A$ that it will rain in city $A$ tomorrow and the boolean valued $X_B$ that the minimum temperature in city $B$ will be below freezing have the same Bernoulli distribution if the probability $p$ happens to be the same. An infinite sequence that alternates $X_A$ and $X_B$ converges in distribution to the Bernoulli distribution with parameter $p$ though the two probability spaces are entirely different. – Jayanth R Varma Jun 24 '25 at 06:25
  • I can buy that convergence in measure need not be topological convergence due to not even being defined on the same probability space. It's subtle but it counts. It seems like a fix, though, is to require that all the measures be on the same space. Would such a requirement address the counterexample? – cgmil Jun 24 '25 at 16:01
  • @cgmil convergence in distrubtion and in measure are not the same thing. The latter is metrizable – Jakobian Jun 24 '25 at 19:43
  • Edited the answer to add another weird problem with convergence in distribution in case of symmetric distributions. – Jayanth R Varma Jun 25 '25 at 11:06