# Convergence in distribution of the characteristic function to the Dirac delta

The question is: does the sequence of characteristic functions $$f_k(x) := \chi_{[-\frac{1}{k}, \frac{1}{k}]}(x)$$ converge in distributional sense to the Dirac delta?

In order to answer I followed this approach, but I fear I'm neglecting something important in my lines:

Firstable, $$f_k\in L^1_{loc}(\mathbb{R})$$, so we can write the action of the associated distribution as $$\langle T_k(x),\psi \rangle= \int_\mathbb{R}\chi_{[-\frac{1}{k}, \frac{1}{k}]}(x)\cdot\psi(x)dx=\int_{-\frac{1}{k}}^\frac{1}{k}\psi(x)dx$$ for every test function $$\psi \in C^\infty_c(\mathbb{R})$$. Then I computed the limit as: $$\displaystyle{\lim_{k\to \infty}\langle T_k(x),\psi\rangle =\lim_{k\to \infty} \int_{-\frac{1}{k}}^\frac{1}{k}{\psi(x)dx} =\lim_{k\to \infty} \int_{-1}^1{\frac{1}{k}\cdot\psi(\frac{y}{k})dy}}$$ and applied the Lebesgue dominated convergence theorem, saying that $$\frac{|\psi(\frac{y}{k})|}{|k|} \le \sup_{x\in [-1,1]}|\psi(x)| <\infty$$ and $$\displaystyle{\lim_{k\to \infty}{\frac{\psi(\frac{y}{k})}{k}} = 0 }$$. Then I deduced that the sequnce $$T_k$$ converges to the distribution associated to the zero function.

Is my proof correct? Any check or observation would be really appreciated.

Your argument is correct. Indeed, characteristic functions $$\chi_\varepsilon$$ of smaller and smaller intervals $$[-\varepsilon,\varepsilon]$$ do not converge to $$\delta$$ (in a distributional sense), but, rather, to $$0$$. Still, the renormalizations to have "total mass" $$1$$, namely, $${1\over 2\varepsilon}\chi_{\varepsilon}$$, do converge to $$\delta$$, by a similar computation.
You are making things too commplicated. $$\psi$$ is a bounded function and if $$|\psi| \leq M$$ we get $$|\int_{-1/k}^{1/k} \psi (x)dx|\leq \frac M {2k} \to 0$$.