Prove or disprove $\lim_{K \to\infty} \lim_{n\to\infty}E(\min(X_n,K))=E(X)$ when $X_n$ surely converges to $X$

Question

The Question:

Let $X, X_1, X_2, ...$ be non-negative random variables, defined jointly on some probability triple $(\Omega,\mathcal F,\mathbf P)$, each having finite expected value. Assume that $\lim_{n\to\infty}X_n(\omega)=X(\omega)$ for all $\omega \in \Omega$. For $n$, $K \in \mathbf N$, let $Y_{n,K}=\min(X_n,K)$. For each of the following statements, either prove it must be true, or provide a counter-example to show it is sometimes false.

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

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

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I don't even know where to start. Thanks for any help.

Answer 1

Hints for (a):

1. Use the dominated convergence theorem to show that $$\lim_{n \to \infty} \mathbb{E}(Y_{n,K}) = \mathbb{E}(\min\{X,K\}).$$
2. It follows from step 1 that (a) is equivalent to $$\lim_{K \to \infty} \mathbb{E}(\min\{X,K\}) = \mathbb{E}(X). \tag{1}$$
3. Apply the monotone convergence theorem to prove $(1)$.

Hints for (b):

1. Use the dominated convergence theorem to show that $$\lim_{K \to \infty} \mathbb{E}(Y_{n,K}) = \mathbb{E}(X_n).$$
2. By step 1, (b) is equivalent to $$\lim_{n \to \infty} \mathbb{E}(X_n) = \mathbb{E}(X). \tag{2}$$
3. Try to find a counterexample to $(2)$ (e.g. a sequence such that $X_n \to 0$ almost surely, but $\mathbb{E}(X_n)=1$ for all $n \in \mathbb{N}$.)