Proof of Smooth Test Function Criterion for Convergence in Distribution

Claim 1. A $C^3$ map is in particular differentiable, hence continuous at every point (directly from the definition of the derivative and limit arithmetic), so $f$ is Borel measurable by the generator criterion of that definition (preimages of open sets are open). Consequently $f\circ Y$ is a random variable, as preimages compose. If $|f|\le M$ pointwise, then $|f\circ Y|\le M$ pointwise; the constant $M$ is a nonnegative simple function with integral $M\cdot P(\Omega)=M$ on a probability space, so by monotonicity (claim 1 of Linearity and Monotonicity of the Lebesgue Integral) $\int|f\circ Y|\,dP\le M<\infty$ and $f\circ Y$ is integrable; its expectation is a real number.

Claim 2, Step 1 (smooth monotone transitions). Let $h:\mathbb{R}\to\mathbb{R}$ be the function $h(v)=\exp(-1/v)$ for $v>0$ and $h(v)=0$ for $v\le 0$, which is a smooth map with $0\le h$ and $h>0$ exactly on $(0,\infty)$, by Step 1 of the proof of Existence of Smooth Bump Functions on Euclidean Space (with the exponential function now the published one; the argument is identical). For real $a<b$ define

$$\eta_{a,b}(x)=\frac{h(b-x)}{h(b-x)+h(x-a)}.$$

The denominator is everywhere positive (for each $x$ at least one of $b-x>0$, $x-a>0$ holds), so $\eta_{a,b}$ is smooth by Products and Quotients of C^k Real-Valued Maps on Euclidean Open Sets Are C^k (quotient with nonvanishing denominator) and the smoothness of the two affine compositions; moreover $0\le\eta_{a,b}\le 1$, $\eta_{a,b}=1$ on $(-\infty,a]$ (there $h(x-a)=0$ and $h(b-x)>0$) and $\eta_{a,b}=0$ on $[b,\infty)$. Every derivative of $\eta_{a,b}$ is continuous and vanishes on $(-\infty,a)\cup(b,\infty)$ (where the function is locally constant), hence is bounded: on the compact interval $[a,b]$ it attains a maximum and minimum by Extreme Value Theorem on a Compact Interval, and it vanishes elsewhere. Therefore $\eta_{a,b}$ is an admissible test function in the sense of the statement.

Claim 2, Step 2 (squeezing the distribution functions). Fix $t\in\mathbb{R}$ at which $F_X$ is continuous, and let $\varepsilon>0$. Consider $\eta^{+}=\eta_{t,t+\varepsilon}$ and $\eta^{-}=\eta_{t-\varepsilon,t}$. Pointwise,

$$\mathbf{1}_{(-\infty,t]}\ \le\ \eta^{+}\ \le\ \mathbf{1}_{(-\infty,t+\varepsilon]},\qquad \mathbf{1}_{(-\infty,t-\varepsilon]}\ \le\ \eta^{-}\ \le\ \mathbf{1}_{(-\infty,t]}.$$

Taking expectations of the compositions with $X_m$ and $X$ and using monotonicity (Linearity and Monotonicity of the Lebesgue Integral) together with $\mathbb{E}[\mathbf{1}_{(-\infty,u]}(Y)]=P(Y\le u)=F_Y(u)$ (Simple Function and Its Integral, Distribution and Cumulative Distribution Function of a Random Variable),

$$F_{X_m}(t)\ \le\ \mathbb{E}[\eta^{+}(X_m)],\qquad \mathbb{E}[\eta^{+}(X)]\ \le\ F_X(t+\varepsilon),$$ $$\mathbb{E}[\eta^{-}(X_m)]\ \le\ F_{X_m}(t),\qquad F_X(t-\varepsilon)\ \le\ \mathbb{E}[\eta^{-}(X)].$$

By hypothesis $\mathbb{E}[\eta^{\pm}(X_m)]\to\mathbb{E}[\eta^{\pm}(X)]$, so with $\liminf$ and $\limsup$ of bounded real sequences as in the proof of Dominated Convergence Theorem,

$$F_X(t-\varepsilon)\ \le\ \liminf_m F_{X_m}(t)\ \le\ \limsup_m F_{X_m}(t)\ \le\ F_X(t+\varepsilon).$$

Letting $\varepsilon\to 0$ and using continuity of $F_X$ at $t$, both outer bounds converge to $F_X(t)$; hence $F_{X_m}(t)\to F_X(t)$. Since $t$ was an arbitrary continuity point, $X_m\to X$ in distribution by Almost Sure Convergence, Convergence in Probability, and Convergence in Distribution. $\blacksquare$