Proof. We prove it under the extra assumption that the moment generating function of \(X_i\) exists near zero; the general proof is the same with the characteristic function in its place. Let \(W_i = (X_i - \mu )/\sigma \), with moment generating function \(M(s)\), and \(S_n = \sqrt n(\bar X_n - \mu )/\sigma = n^{-1/2}\sum W_i\). By independence, \[ M_{S_n}(t) = \bigl [M(t/\sqrt n)\bigr ]^n . \] Since \(E(W) = 0\) and \(E(W^2) = 1\), Taylor’s theorem gives \(M(s) = 1 + \frac 12s^2 + r(s)\) with \(r(s)/s^2 \to 0\) as \(s \to 0\). Hence \[ \log M_{S_n}(t) = n\log \Bigl (1 + \frac {t^2}{2n} + r(t/\sqrt n)\Bigr ) \longrightarrow \frac {t^2}{2}, \] because \(n\,r(t/\sqrt n) = t^2\,r(t/\sqrt n)/(t/\sqrt n)^2 \to 0\) and \(n\log (1 + c_n/n) \to c\) whenever \(c_n \to c\). The limit \(e^{t^2/2}\) is the moment generating function of \(N(0, 1)\), and convergence of moment generating functions on a neighbourhood of zero implies convergence in distribution (Curtiss’s continuity theorem). Multiplying by \(\sigma \) gives the result. □

Proof. We prove the first statement; the others follow by similar arguments. Let \(F\) be the distribution function of \(X\), and let \(x\) be such that \(F\) is continuous at \(x - c\). For \(\varepsilon > 0\) such that \(F\) is continuous at \(x - c \pm \varepsilon \), \begin {align*} P(X_n + Y_n \leq x) &\leq P(X_n \leq x - c + \varepsilon ) + P(|Y_n - c| > \varepsilon ),\\ P(X_n + Y_n \leq x) &\geq P(X_n \leq x - c - \varepsilon ) - P(|Y_n - c| > \varepsilon ) . \end {align*}

Letting \(n \to \infty \), the limit of \(P(X_n + Y_n \leq x)\) lies between \(F(x - c - \varepsilon )\) and \(F(x - c + \varepsilon )\). Letting \(\varepsilon \to 0\) through continuity points, both bounds tend to \(F(x - c) = P(X + c \leq x)\). □

Proof. First, \(T_n \xrightarrow {p} \theta \): by Slutsky, \(T_n - \theta = n^{-1/2}\cdot \sqrt n(T_n - \theta ) \xrightarrow {d} 0 \cdot N(0, v) = 0\), and convergence in distribution to a constant is convergence in probability to it. Define \(h(t) = \bigl (g(t) - g(\theta )\bigr )/(t - \theta )\) for \(t \neq \theta \) and \(h(\theta ) = g'(\theta )\). Differentiability makes \(h\) continuous at \(\theta \), so \(h(T_n) \xrightarrow {p} g'(\theta )\) by Theorem 4.2.5. Then \[ \sqrt n\bigl (g(T_n) - g(\theta )\bigr ) = h(T_n)\cdot \sqrt n(T_n - \theta ) \xrightarrow {d} g'(\theta )\,N(0, v) \] by Slutsky’s theorem. □