Usbrth

Specifically, this is about Lemma 5.1 of A. M. Vershik's "Statistical mechanics of combinatorial partitions and their limit shapes," but I will ask my question in some generality.

Let $mathcalF(x):= sum_n geq 1 Q_n x^n$ for $x in [0,1)$ and $Q_n >0$.

Let $mu_n_n geq 1$ and $mu_x_x in [0,1)$ be, respectively, a discrete and continuous family of probability measures on the same space, where the $mu_x$ are defined by $$mu_x := mathcalF(x)^-1sum_n geq 1 x^n Q_n mu_n.$$ This is called a convex combination because the coefficients sum to 1.

Vershik states that, because the $mu_x$ are convex combinations, it is obvious that if the $mu_n$ have a weak limit, then the $mu_x$ have the same limit as $x to 1^-$.

Question: Why is the above true?

I can at least guess why it should maybe be true if we assume $mathcalF(1)= infty$, so $mathcalF(1)^-1=0$ (which is the case for the $mathcalF$'s in his paper).

Since $$mu_x = mathcalF(x)^-1sum_n = 1^N x^n Q_n mu_n+ mathcalF(x)^-1sum_n > N x^n Q_n mu_n,$$ we have $$ lim_x to 1^- mu_x = lim_x to 1^- mathcalF(x)^-1sum_n > N x^n Q_n mu_n.$$ Now choosing $N$ so that the $mu_n$'s are "close" to the weak limit $mu$, the above is "roughly" $$left(lim_x to 1^- mathcalF(x)^-1sum_n > N x^n Q_n right) mu = left(lim_x to 1^- mathcalF(x)^-1 left(mathcalF(x) - sum_n=1^N x^nQ_n right) right) mu = mu.$$

Am I on the right track?

Yes, you are on the right track; you just need to apply the definition of weak convergence. You do need $mathcal F(x)toinfty.$

A sequence of probability measures $nu_n$ converges weakly to $nu$ iff for each bounded continuous $f$ and each $epsilon>0,$ there exists $N$ such that $|int f dnu_n-int f dnu|$ for $n>N.$ We can write this as: $nu_n-nuin E$ for all sufficiently large $n,$ where $E$ is the set of finite signed measures $tau$ with $|int f dtau|leqepsilon.$
Fix $f$ and $epsilon$ and let $E$ be as I just defined.

We are given that $mu_n$ converges weakly to $mu.$ Applying the definition of weak convergence with $f$ and $tfrac12epsilon,$ there exists $N$ such that for all $n>N$ we have $mu_n-muintfrac12 E.$ In particular, $$mathcal F(x)^-1sum_n>Nx^nQ_n(mu_n-mu)intfrac12 E$$ because this is an infinitary convex combination of elements of the closed convex set $tfrac12 E.$
Since $mathcal F(x)toinfty,$ as $xto 1^-$ we have $$mathcal F(x)^-1sum_n=1^N x^nQ_n(mu_n-mu) in tfrac 12 E$$
because each $Q_n(mu_n-mu)$ lies in $tfrac12Nmathcal F(x)E$ for sufficiently large $mathcal F(x).$ Summing gives
$$mu_x-mu=mathcal F(x)^-1sum_n=1^infty x^nQ_n(mu_n-mu) in E$$
which is the definition of $mu_x$ converging weakly to $mu.$

You might find it clearer to do everything in terms of real numbers - just replace $mu_x,mu_n,mu$ by $int f mu_x$ etc, and replace $E$ by the interval $[-epsilon,epsilon].$