Sequence of random variables $\left\{X_n\right\}_{n=1}^\infty$ converge to random variable $X$ in Kolmogorov distance if $$\lim\limits_{n\to \infty}\left(\sup\limits_{x\in\mathbb{R}}|F_n(x)-G(x)|\right)=0,$$ where $F_n$ is distributive function of random variable $X_n$ and $G$ is distributive function of random variable $X$.
I know that convergence in Kolmogorov distance implies convergence in distribution (converge weakly, or converge in law). Also I know that convergence in distribution does not imply convergence in Kolmogorov distance. Somewhere I have been reading that convergence in distribution imply convergence in Kolmogorov distance when distribution function $G$ is absolutely continuous.
But I cannot find example when convergence in distribution is hold but convergence in Kolmogorov distance is not. Any help will be appreciated. Thanks
This is not really about convergence of random variables, but rather convergence in functions. Convergence in distribution is essentially point-wise convergence, whereas convergence in Kolmogorov distance is uniform convergence.
For the example, consider $G(x)$ to be the Heaviside step function (the CDF of the zero random variable), and let $F_{n}(x)$ be the CDF of the normal distribution with mean zero and variance $1/n$, which essentially looks like a logistic curve.
Then $F_{n}(x)$ converges pointwise to $G(x)$. This is clear for all $x \not = 0$. For $x = 0$, we can either define $G(0) = 1/2$ or just ignore it, since we don't actually need pointwise convergence for points where $G$ is discontinuous.
In contrast, not matter how large we choose $n$,
$$ \sup_{x} |F_{n}(x) - G(x)| = 1/2$$
since for $x$ slightly greater than zero, $G(x) = 1$ yet $F_{n}(x) \approx 1/2$. Therefore $F_{n}(x)$ does not converge to $G(x)$ in Kolmogorov distance.