I have been stuck with the following problem:
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of integrable functions in $\mathbb{R}^N$ that satisfies: $$\int_{\mathbb{R}^N}f_n(x)dx = 1 \quad \forall n \in \mathbb{N}$$ And given a compact subset $K \subset \mathbb{R}^N$ such that $0 \notin K$, $$\int_{K}f_n(x)dx \xrightarrow{n \to \infty} 0$$ Then the corresponding distribution sequence $(T_{f_n})_{n \in \mathbb{N}}$ converges to the Dirac Delta distribution.
Here's my attempt: We have to prove that, given a test function $\varphi$, $\forall \epsilon > 0 \exists n_0 \in \mathbb{N}$ such that if $n \geq n_0$: $$\left|\int_{\mathbb{R}^N}f_n(x)\varphi(x)dx - \varphi(0)\right| < \epsilon$$ Fix $\varphi \in \mathcal{D}$, the set of test functions with compact support. Let $K = supp(\varphi)$, $0 < \rho < 1$, and $\overline{B}_\rho = \{x \in \mathbb{R}^N: |x| < \rho\}$. Then:
$$\left|\int_{\mathbb{R}^N}f_n(x)\varphi(x)dx - \varphi(0)\right| = \left|\int_{\mathbb{R}^N}f_n(x)\varphi(x)dx - \int_{\mathbb{R}^N}f_n(x)\varphi(0)\right|$$$$ = \left|\int_{\mathbb{R}^N}f_n(x)(\varphi(x) - \varphi(0))dx\right|$$ We can "split" $\mathbb{R}^N = \overline{B}_\rho\cup(K - \overline{B}_\rho)\cup(\mathbb{R}^N - K)$. Then we have:
$$\left|\int_{\overline{B}_\rho}f_n(x)(\varphi(x) - \varphi(0))dx + \int_{K - \overline{B}_\rho}f_n(x)(\varphi(x) - \varphi(0))dx + \int_{\mathbb{R}^N - K}f_n(x)(\varphi(x) - \varphi(0))dx\right|$$
I know how to make the first and second terms small, but I can't seem to get anything from the third. All I know is that we are beyond the support of $\varphi$, then it is 0 there. But I am left with: $$\int_{\mathbb{R}^N - K}f_n(x)(- \varphi(0))dx$$ Which I have no clue how to find a bound.
I need some help.
Let $$f_n(x) = \begin{cases} 1,&x\in [n,n+1]\\ 0, & \text{otherwise} \end{cases}$$
Clearly, $\forall n\, \int_\Bbb R f_n=1$ and on any compact $K\subset \Bbb R$ we have $\int_K f_n \to 0$ as $n\to\infty$ (because $K$ is bounded).
However, for obvious reasons, $f_n$ does not converge to $\delta_0$.