# How to (dis)prove $\lim_{k\to\infty}\lim_{n\to\infty}E(Y_{n,K}) = E(\min(X_n, K))$?

Here is the problem:

Given $$X, X_1, X_2, \ldots$$, non-negative random variable with finite expectation and $$X_n \to X$$ pointwise and $$Y_{n,K} = \min(X_n,K)$$, we are asked to see if

a) $$\lim_{K \to \infty} \lim_{n \to \infty} E(Y_{n,K}) = E(X)$$

and

b) $$\lim_{n \to \infty} \lim_{K \to \infty} E(Y_{n,K}) = E(X)$$.

For the first problem, here is my attempt.

For fixed $$K$$, $$\min(X_n, K) \leq K$$. So by the dominated convergence theorem, $$\lim_{n \to \infty} E(Y_{n,K}) = E(\lim_{n \to \infty} \min(X_n,K)) = E(\min(X,K)).$$

And hence $$\lim_{K \to \infty} \lim_{n \to \infty} E(Y_{n,K}) = E(X).$$

The second part, for fixed $$n$$, $$Y_{n,K} \leq X_n$$, with $$X_n$$ having a finite expectation. So $$\lim_{K \to \infty} E(Y_{n,K}) = E(X_n),$$ and so $$\lim_{n \to \infty} \lim_{K \to \infty} E(Y_{n,K}) = E(X).$$

Is this correct? Seems like I missed something.

Source: A First Course in Rigorous Probability Theory by Jeffrey Rosenthal.

• The random variables $X_n:(\mathbb R, \mathcal{B}(\mathbb{R}), \lambda)\to\mathbb R$ (Lebesgue measure) given by $X(x) = nI(0\lt x \le 1/n)$ are worth contemplating. Each has unit expectation but their pointwise limit is the zero function, with zero expectation.
– whuber
Dec 7, 2022 at 4:38
• @whuber was the part $\lim_{n \to \infty} \min(X_n, K) = \min(X,K)$ wrong?
– Phil
Dec 7, 2022 at 4:48