Let's break down and **prove** the statement step by step: --- ### **Restate the Problem** Given: - \( X_n \xrightarrow{d} X \) (i.e., \( X_n \) converges in distribution to \( X \)). - \( f \) is continuous on bounded interval \([a, b]\), and defined at all continuity points of \( F_X \). - Need to **show:** \[ \mathbb{E}[f(X_n)] \to \mathbb{E}[f(X)] \] as \( n \to \infty \). **Hint:** You may use the approximation theorem for simple functions. --- ## **Step 1: Use Convergence in Distribution** By definition, \( X_n \xrightarrow{d} X \) means for every bounded and continuous function \( g \), \[ \mathbb{E}[g(X_n)] \to \mathbb{E}[g(X)] \] as \( n \to \infty \). ### **Key Observation** - If \( f \) is bounded and continuous, this is already true by the **Portmanteau theorem**. --- ## **Step 2: Applying the Portmanteau Theorem** **Portmanteau Theorem:** If \( X_n \xrightarrow{d} X \) and \( f \) is bounded and continuous, \[ \mathbb{E}[f(X_n)] \to \mathbb{E}[f(X)] \] So, the result is immediate **if \( f \) is bounded and continuous**. --- ## **Step 3: Approximation by Simple Functions (as per Hint)** If \( f \) is not necessarily simple, but continuous and bounded, we can **approximate \( f \) uniformly by simple functions**. Let \( s \) be a simple function such that for every \( \epsilon > 0 \), \[ |f(x) - s(x)| < \epsilon \quad \forall x \in [a, b] \] --- ## **Step 4: Bounding the Expectation Difference** Let’s break down the error: \[ |\mathbb{E}[f(X_n)] - \mathbb{E}[f(X)]| \leq |\mathbb{E}[f(X_n) - s(X_n)]| + |\mathbb{E}[s(X_n) - s(X)]| + |\mathbb{E}[s(X) - f(X)]| \] But, - \( |\mathbb{E}[f(X_n) - s(X_n)]| \leq \mathbb{E}|f(X_n) - s(X_n)| < \epsilon \) - \( |\mathbb{E}[s(X) - f(X)]| < \epsilon \) (by approximation) The middle term, \( |\mathbb{E}[s(X_n) - s(X)]| \), goes to zero as \( n \to \infty \) **because \( s \) is simple and bounded** (and thus measurable, and the Portmanteau theorem applies for simple functions). --- ## **Step 5: Conclusion** Since \( \epsilon > 0 \) is arbitrary, \[ \lim_{n \to \infty} \mathbb{E}[f(X_n)] = \mathbb{E}[f(X)] \] --- ## **Final Answer (Summary):** ### **Step-by-step summary:** 1. \( X_n \xrightarrow{d} X \) means expectations of bounded, continuous functions converge. 2. Approximate \( f \) uniformly by simple functions \( s \). 3. Use triangle inequality to bound the difference by three terms, each of which can be made arbitrarily small. 4. By the Portmanteau theorem and the approximation, the total difference goes to zero. ### **Therefore:** \[ \boxed{ \lim_{n \to \infty} \mathbb{E}[f(X_n)] = \mathbb{E}[f(X)] } \] **This is a classical result, often called the "Continuous Mapping Theorem for Expectations" for bounded continuous functions.**