Show that $ Χ_n \rightarrow X$ (Challenging Problem?) Suppose that $X_n$ assumes as values $γ_n + kδ_n$ , $k= 0 $, $\pm 1 $, ..., where $δ_n >0$ .
Suppose that $δ_n \rightarrow 0$ and that , if $k_n$ is an integer varying with $n$ in such a way 
that $γ_n + k_n δ_n  \rightarrow x $, then $P[ X_n = γ_n + k_n δ_n]{δ_n}^{-1} \rightarrow f(x)$,
where $f$ is the denstiy of a random variable $X$ . Show that  $    Χ_n \rightarrow X$ .
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Attempt : 
We construct a probability space and random
variables $U , X_1 , X_2 , X_3$ .. defined on that space with each $X_n$ having the distribution specified in the problem and with $U$ ~ Uniform $(0,1)$ independent of $X_n$ for all $n$.
Let $Y_n = X_n + δ_nU$ 
How can we:
(i)Calculate  the density $f_n$ of $Y_n$ and then 
(ii)Show that $f_n(x) \rightarrow  f(x)$ as $n \rightarrow  \infty$
 A: We have $\lim_n \delta_n^{-1}P(X_n=\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)=f(x),$ from the above assumption. 
So let $h$ be a continuous and bounded function. $M:=\sup_x |h(x)|.$
We have $$|\int_{\mathbb{R}}h(x)dP_{X_n}(x)-\int_{\mathbb{R}}h(x)f(x)dx|=|\sum_{k \in \mathbb{Z}}h(\gamma_n+k\delta_n)P(X_n=\gamma_n+k\delta_n)-\int_{\mathbb{R}}h(x)f(x)dx|$$
$$=|\frac{1}{\delta_n}\sum_{k \in \mathbb{Z}}\int_{\gamma_n+k\delta_n}^{\gamma_n+(k+1)\delta_n}h(\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)P(X_n=\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)dx-\int_{\mathbb{R}}h(x)f(x)dx|$$
$$=|\int_{\mathbb{R}}\frac{1}{\delta_n}h(\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)P(X_n=\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)dx-\int_{\mathbb{R}}h(x)f(x)dx|$$
$$\leq \int_{\mathbb{R}}|h(\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)(\frac{1}{\delta_n}P(X_n=\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)-f(x))|dx+\int_{\mathbb{R}}f(x)|h(\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)-h(x)|dx$$
$$\leq M \int_{\mathbb{R}}|\frac{1}{\delta_n}P(X_n=\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)-f(x)|dx+\int_{\mathbb{R}}f(x)|h(\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)-h(x)|dx$$
using Scheffé's lemma the first term converges to $0$ since $$\int_{\mathbb{R}}\frac{1}{\delta_n}P(X_n=\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)dx=1=\int_{\mathbb{R}}f(x)dx$$ and using the continuity of $h,$ we have $\lim_n f(x)|h(\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)-h(x)|=0$ and since $f(x)|h(\gamma_n+\left \lfloor{\frac{x-\gamma_n}{\delta_n}}\right \rfloor\delta_n)-h(x)|\leq 2Mf(x),$ so using the dominated convergence theorem the second term converges to $0,$ and then $X_n$ converges in distribution to $X$ such that $$P_X(B)=\int_{B}f(x)dx.$$ 
