convergence of $X_n = n\chi_{(0,1/n]}$ in distribution .vs weakly

From what I learnt, convergence in distribution, where $X$ is random variable, $$F(X_n<x) \rightarrow F(X<x)$$ is equivalent to weak convergence, $$\mathbb{E}[X_ng] \rightarrow \mathbb{E}[Xg],~\forall g\in L^\infty$$

Now, suppose we are working on uniform distribution $U[0,1]$, and we have a random variable $X_n = n\chi_{(0,1/n]} : [0,1]\rightarrow \mathbb{R}$ which convergence to $X =0$ point wisely, hence $X_n\xrightarrow{a.s.}X$. This implies $X_n\xrightarrow{p}X$ and $X_n\xrightarrow{d}X$. However, if we look at the definition of weak convergence, let $g = \chi_{[0,1]}$ $$\int_{[0,1]} X_ng = \int_{[0,1]} n\chi_{(0,1/n]} =1$$ when $$\int_{[0,1]} Xg =0$$ Hence, $X_n$ does not converge to $X$ weakly.

Since there is an theorem stating that convergence in distribution is equivalent to weak convergence, I assume I made a mistake somewhere. Please help me find.

Weak convergence means $\lim_{n \to \infty}(E(f(X_n))=E(f(X))$ for $f$ bounded and $f \in \mathcal{C}^\infty$
I think the problem is that the $f$ function has to be bounded, in your case as you are taking the identity function $f(x)=x$ this isn't fulfilled.
The thing here is that the random variable you are proposing doesn't satisfy the assumptions of Lebesgue's dominated convergence theorem nor those of the monotone convergence theorem hence $\lim_{n \to \infty} E(X_n) \neq E(\lim_{n \to \infty}X_n)$
• My $f$ is $\chi_{0,1}$ which is bounded by 1. Looks like your definition of weak definition requires $f$ to be smooth. Do you know any reason why definition of weak convergence differs between probability theory and measure theory? – user1292919 Nov 8 '17 at 21:51
• You are taking $g$ as $\chi_{0,1}$ that is equal to taking $f$ as the identity since $\chi_{0,1}=1$ over all relevant values, and note that I am using function composition instead of multiplication. What's the name of the book you are following? – Daniel Ordoñez Nov 9 '17 at 19:03