How to show convergence in distribution

Question

Let $([0,1],B,\lambda)$ (B Borel Sigma-algebra) and $\lambda$ the Lebesgue measure. I want to show that this sequence converges in distribution. $$X_n(\omega)= \left(\begin{matrix} 1, & \mbox{iff} \frac{m}{2^k} \le \omega \le \frac{m+1}{2^k}\\ 0, & \mbox{otherwise.} \end{matrix}\right)$$ This is obvious, cause it converges in $L^p$ and it converges in measure( this was the way I did it). Despite, I noticed that besides some discrete examples, there are very few examples where you show the convergence in distribution directly. I know that convergence in distribution can be shown via the convergence of the distribution function where it is continuous, by the pointwise convergence of the characteristic function and by the convergence of $E(f \circ X_n) \rightarrow E(f\circ X)$ where $f$ is an arbitrary bounded and continuous function. Honestly, I don't know what the best strategy of the three ones possible would be, cause they all sound difficult. Therefore, I was wondering whether any user experienced with this could explain to me how he would show it or which of the three possible ones he would use?

Answer 1

If we want to check directly the convergence in distribution of $(X_n)$ to $0$, then we write for $f\colon [0,1]\to\mathbf R$ continuous $$\int_{[0,1]}f(X_n(t))\lambda(dt)=\int_{m/2^k}^{(m+1)/2^k}f(x)dx. $$ This gives that $$\left|\int_{[0,1]}f(X_{2^k+m}(t))\lambda(dt)\right|\leqslant 2^{-k}\sup_{x\in [0,1]}|f(x)|, $$ hence $$\left|\int_{[0,1]}f(X_{n}(t))\lambda(dt)\right|\leqslant 2^{-\lfloor \log _2n\rfloor}\sup_{x\in [0,1]}|f(x)|$$ and the conclusion follows.