# Intuition: Portmanteau-Theorem

An important theorem in probability theory about weak convergence of measures is the Portmanteau-Theorem.

Why should it be true - intuitively - though?

EDIT: Our version of Portmanteau's Theorem is: The following statements are equivalent

1. $$\mu_n {\to} \mu$$ weakly
2. $$\int f \ d \mu_n \to \int f \ d\mu$$ for all uniformly continuous and bounded $$f:S \to \mathbb{R}$$
3. $$\limsup_{n \to \infty} \mu_n(F) \leq \mu(F)$$ for all measurable closed subsets
4. $$\liminf_{n \to \infty} \mu_n(U) \geq \mu(U)$$ for all measurable open subsets $$U$$
5. $$\lim_{n \to \infty}\mu_n(A) = \mu(A)$$ for all measurable $$A$$ with $$\mu(\partial A) = 0$$
• It may help to state your version of the theorem, and why you think it is unintuitive. It involves probability measures that behave similarly and so many people would have the opposite perspective: They would find it to be intuitive. As I recall, the main concept is the connection between expectations and probabilities via indicator functions $$P[X\leq x] = E[1_{\{X\leq x\}}]$$ – Michael May 26 at 19:08
• I edited my post and stated our Portmanteau's Theorem. I don't quite see what the latter conditions have to do with weak convergence intuitively. How does your connection between expectations and probabilities play a role? Thanks! – Qi Zhu May 26 at 19:16
• Your version involves integrals and measures (generalizations of "expectations and probabilities") and so the equation I wrote above becomes, for all measurable sets $A$, $$\mu(A) = \int 1_{\{x \in A\}}d\mu$$ Now the indicator function $1_{\{x \in A\}}$ is bounded but is not continuous. However it can easily be approximated by continuous functions. So if we get a desirable convergence for all bounded continuous functions, we also get a desirable convergence for the particular ones that are good approximations of indicator functions. – Michael May 26 at 19:21
• Hm yes, that's the also the essential idea in our proof of the theorem. But are these latter conclusions (perhaps especially the last) "intuitively" clear with that interpretation without going through computational work? I.e. after noting that, could one immediately guess that the last equivalence should be equivalent? (I apologize for these vague questions...) – Qi Zhu May 26 at 19:27
• Indeed, after an hour or two of thinking, one can immediately guess the result. =) – Michael May 26 at 19:28

Let's momentarily ignore 3. and 4..

Hopefully you agree that any notion of convergence of distributions $$\mu_n$$ to a distribution $$\mu$$ would imply $$\mu_n(A) \rightarrow \mu(A)$$ for an appropriate class of $$A$$. In this case the equivalence of 1. and 5. should be natural. The requirement that $$\mu(\partial A) = 0$$ allows us to say $$\delta_{1/n} \Rightarrow \delta_0$$, where $$\delta_x(dx)$$ is the measure $$\delta_x(A) = 1$$ if $$x \in A$$ and $$0$$ otherwise.

Regarding 2., you should poke this definition for yourself a little bit. Compare with the definition of weak-$$*$$ convergence on the Banach space $$X = C(S)$$ for $$S = [0, 1]$$ and $$S = [0, \infty)$$ to get a feeling of this "weak" formulation of convergence. Keep in mind that the space of probability measures $$\mathcal{P}(S)$$ on a Polish space $$S$$ is not a vector space as it is not closed under addition. An equivalent statement of 2. is $$\int f d\mu_n \rightarrow \int f d\mu$$ for $$f$$ only bounded and continuous ($$f\in C_b(S)$$), and one can even relax this to $$f \in C_b(S)$$ which additionally vanish at infinity (aka $$f \in C_0(S)$$) under the assumption that $$\mu$$ is itself a probability measure (consider $$\mu_n = \delta_n$$). These puzzles are, as others have noted, up to you to think about.

If Portmanteau were just 1., 2. and 5., I suspect you would not have posted this question. What is confusing is that you can get away with the apparently "weaker" inequalities 3. and 4., and I interpret your question as "why am I not actually giving anything up?".

First, note that 3. and 4. are directly equivalent, simply by taking complements. Then, note that 3. and 4. together imply for any set $$A$$, $$\mu(A^\circ) \leq \liminf_n \mu_n(A) \leq \limsup_n \mu_n(A) \leq \mu(\bar{A}),$$ where $$A^\circ$$ and $$\bar{A}$$ are the interior and closure of $$A$$, respectively. In particular, if $$\mu(\bar{A}) = \mu(A^\circ)$$, then $$\mu_n(A) \rightarrow \mu(A)$$. If you recall that $$\bar{A} = A^\circ \cup \partial A$$, you recover 5.

That is the "formal" explanation, but I would say you should look deeper at semicontinuity definitions. Consider, for instance, that a lower semicontinuous function has closed level sets and acheives its minimum on compact sets. These definitions are strong enough to guarantee some notion of convergence, but flexible enough that they extend to general contexts. Take a look at $$\Gamma$$-convergence or the Large deviations principle.

My point is these inequalities are useful to keep in mind in their own right, and the things to think about are how you would approximate the indicator function of an open set or a closed set. This is how the Portmanteau theorem is usually proved, and the best reference for weak convergence in general is Billingsley's book Convergence of Probability Measures.

• Excellent answer, thank you! – Qi Zhu May 27 at 15:13

This is a subjective question; I assume all answers are subjective.

My take is, weak convergence of measures is such a natural concept that is has been rediscovered many times, and formalized slightly differently by each rediscover. The Portmanteau theorem is says, in effect, all these definitions are equivalent.

• Surely mathematical intuition is somewhat subjective but still IMHO the most important thing in mathematics. Perhaps it would then help to ask what makes weak convergence natural and how people would find that these are equivalent. There should always be a deeper reason as to why natural concepts are equivalent - and that's what I'm looking for. – Qi Zhu May 26 at 19:19
• I don't see how to help you. Keep studying, maybe things will seem more intuitive later on. Vol. 2 of Bogochov's measure theory has a thorough chapter on weak convergence, plus (in his historical notes) a diatribe against the "nonsensical name" Portmanteau theorem. – kimchi lover May 26 at 19:27
• Mhm, sure, of course things will get more intuitive later on. Thank you for the reference of Bogochov's book, I like historical notes. :) – Qi Zhu May 26 at 19:29