Weak convergence and the Portmanteau theorem
Translate weak convergence into usable statements about events while respecting boundaries, atoms, and unbounded observables.
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The question for today
What you’ll understand
- Use bounded continuous tests to define weak convergence of probability measures.
- Apply the closed-set, open-set, and continuity-set forms of the Portmanteau theorem.
- Recognize why weak convergence does not guarantee every event probability or unbounded moment converges.
When an approximation changes the sample space, comparing outcomes point by point may have no meaning. Weak convergence instead asks whether the probability laws become indistinguishable to bounded continuous measurements. This is a natural language for replacing a finely spaced discrete distribution by a continuous one. The important qualification is continuity: a measurement that jumps exactly at an atom can detect a difference that persists even as the atoms move an arbitrarily small distance. The Portmanteau theorem describes precisely which event comparisons remain valid.
Test the law using continuous measurements
Let μn and μ be probability measures on a metric space, with its Borel sets. Weak convergence means that integrating any bounded continuous real function against μn approaches its integral against μ. Boundedness prevents a tiny amount of probability at an enormous value from dominating the test. Continuity prevents the test from magnifying an arbitrarily small movement across a sharp threshold. On the real line this is equivalent to convergence of cumulative distribution functions at every continuity point of the limiting distribution function.
For a deterministic point mass at 1/n, bounded continuous tests tend to their value at zero, so the laws converge weakly to a point mass at zero. Yet the probability of the singleton containing zero is always zero before the limit and one at the limit. Weak convergence did not fail; the indicator of that singleton is discontinuous at the place where the limit concentrates all its mass. Event probabilities therefore need a boundary condition.
Open and closed sets give one-sided guarantees
The Portmanteau theorem states equivalent criteria for weak convergence. For every closed set, the upper limit of its probabilities is at most its limiting-law probability. For every open set, the lower limit is at least its limiting-law probability. A closed set can acquire mass when nearby points approach its boundary, whereas an open set can lose boundary mass. These one-sided inequalities are oriented differently for a reason; reversing them would already fail for moving point masses.
If a Borel event A has boundary of μ-measure zero, applying the two inequalities to its interior and closure traps its probabilities between two quantities tending to the same value. Then μn(A) tends to μ(A). Such an event is called a continuity set of the limiting measure. The condition is sufficient, not necessary for a particular sequence. On the real line, a threshold event is safe when the limiting law has no atom at the threshold.
Know which transformations and moments survive
A continuous map preserves weak convergence after pushing the measures forward. More generally, a measurable map works when its discontinuity set has zero probability under the limiting law. This is the continuous mapping theorem. Squaring a real random variable is continuous everywhere; thresholding it is discontinuous at the threshold. The distinction explains why a smooth change of coordinates is safe while a decision rule near a limiting atom needs special care.
Weak convergence alone does not control unbounded expectations. A variable equal to n² with probability 1/n and zero otherwise converges in probability, hence in distribution, to zero; its expectation is n. Additional tail control, often uniform integrability, is needed to pass such moments through a weak limit. Tightness is a different idea: it prevents probability mass escaping every compact region and provides convergent subsequences under suitable space assumptions. It does not alone bound the contribution of very large values to an expectation.
Work it through
Check a limit and a threshold separately
Let μn put probability one half at 1/n and one half at 2 + 1/n. Identify its weak limit, evaluate one bounded continuous test in the limit, and compare the events (−0.5,1.5) and (−∞,0].
For any bounded continuous f, the integral is [f(1/n) + f(2 + 1/n)]/2, which tends to [f(0) + f(2)]/2. Thus the limiting law puts half its mass at zero and half at two.
For f(x) = x/(1 + |x|), the limiting integral is [0 + 2/3]/2 = 1/3. The event (−0.5,1.5) has boundary points carrying no limiting mass and probability one half for every positive n.
The event (−∞,0] has μn-probability zero, but limiting probability one half. Its boundary contains zero, which is an atom of the limiting law. The closed-set inequality permits this increase.
The laws converge weakly, and the continuity-set event behaves as expected. A threshold located at a limiting atom can have nonconvergent event probabilities without contradicting weak convergence.
The living notebook
Explore the model
Interactive lab · probability
Interpret concentration through the right coordinate
Set p = 0.25 and compare n = 4, 16, and 30. Interpret a bar at k as sample proportion k/n. Predict that the proportion mean stays at 0.25 while its variance p(1−p)/n decreases.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe lab displays exact binomial count laws, not a weak-limit solver or a transformed proportion axis. The finite bars illustrate concentration toward a point mass after rescaling; they cannot establish Portmanteau inequalities or settle events whose boundary contains that limiting point.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let Un be uniform on {1/n,2/n,…,1}. Show that Un converges weakly to a uniform variable on [0,1], but the probabilities of the set of rational numbers do not converge to the limiting-law probability.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Couple Un to U by rounding nU upward and dividing by n. Then compare the probability of a countable set under the two laws.
Compare with the solution
Take U uniform on (0,1) and define Un = ceil(nU)/n. Each of the n grid values has probability 1/n, and |Un − U| ≤ 1/n, so Un tends almost surely to U and therefore converges weakly. Every grid value is rational, giving rational-set probability one for every n. A uniform continuous law assigns probability zero to each singleton and hence to the countable set of rationals. Its boundary is all of the real line, which has limiting probability one; the continuity-set condition fails exactly where needed.
Keep the idea
Take it with you
- Weak convergence is convergence against bounded continuous measurements.
- Event probabilities converge automatically at sets whose limiting boundary mass is zero.
- Unbounded moments and discontinuous decisions need hypotheses beyond weak convergence.
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