Using the definition of continuity, let $\varepsilon > 0$, then ther exists $\delta >0$ s.t. $\lvert X_n - X \rvert \leq \delta \implies \lvert f(X_n) - f(X) \rvert \leq \varepsilon$ or equivalently (taking the negation of previous statement), $\lvert f(X_n) - f(X) \rvert > \varepsilon \implies \lvert X_n - X \rvert > \delta$. Then by inclusion of events, $\mathbb{P}(\lvert f(X_n) - f(X) \rvert > \varepsilon) \leq \mathbb{P}(\lvert X_n - X \rvert > \delta)$ and the last quantity goes to $0$ as $n\to \infty$ by definition of convergence in probability.
Hint: Notice that $\mathbf{1}_{(x,\infty)}(y) = \begin{cases} 1, & y \in (x,\infty) \\ 0, & y \in (-\infty,x] \end{cases}$. So, if we want to change this to be a function of $x$ instead of $y$, we see that
\begin{align*}
\mathbf{1}_{(x,\infty)}(y) &= \begin{cases} 1, & y \in (x,\infty) \\ 0, & y \in (-\infty,x] \end{cases} \\ &= \begin{cases} 1, & x \in (-\infty,y) \\ 0, & x \in [y,\infty) \end{cases} \\
&= \mathbf{1}_{(-\infty,y)}(x).
\end{align*}
Therefore, the product becomes $\mathbf{1}_{(0,\infty)}(x) \mathbf{1}_{(x,\infty)}(y) = \mathbf{1}_{(0,\infty)}(x) \mathbf{1}_{(-\infty,y)}(x) = \mathbf{1}_{(0,\infty) \cap (-\infty,y)} (x)$.
Best Answer
Consider the second-order central difference of a function $f$: $$ \delta_h^2[f](x)=f(x-2h)+f(x+2h)-2f(x). $$ Then $$ \delta_h^2[\bar{\mu}](t)=\int_{\mathbb{R}}e^{itx}(2i\sin xh)^2\, d\mu(x), $$ and $$ \left|\frac{\delta_h^2[\bar{\mu}](0)}{(2h)^2}\right|=\int_{\mathbb{R}}\left(\frac{\sin xh}{h}\right)^2\, d\mu(x). $$ For each $N>0$, $$ |\bar{\mu}''(0)|=\lim_{h\to 0}\left|\frac{\delta_h^2[\bar{\mu}](0)}{(2h)^2}\right|\ge \lim_{h\to 0}\int_{-N}^N\left(\frac{\sin xh}{h}\right)^2\, d\mu(x)=\int_{-N}^N x^2\,d\mu(x), $$ which implies the result.