Why Dirac's Delta is not an ordinary function? Given the following definition of the Dirac's Delta:
$$\delta: \mathcal{D}(\mathbb{R}^n) \to \mathbb{R}: \varphi \mapsto \langle \delta,\varphi \rangle = \varphi(\mathbf{0})$$
where $\mathcal{D}(\mathbb{R}^n)$ is the space of bump functions over $\mathbb{R}^n$, i.e. the space of smooth functions
$$C_0^\infty(\Omega) := \{ f \in C^\infty(\Omega): \mathrm{supp}(f) \ \mathrm{is \ compact} \}$$
with a suitable topology, and given the following association between an ordinary function $u \in L^1_{loc}(\Omega) $ and the corresponding distribution $I_u \in \mathcal{D}'(\Omega)$
$$ I_u: \mathcal{D}(\Omega) \to \mathbb{R}^n : \varphi \mapsto \langle I_u,\varphi \rangle =\int_\Omega u\varphi $$
how to prove there is not $u \in L^1_{loc}(\mathbb{R}^n)$ such that $I_u = \delta$?
Edit. Or, how to prove that if such $u$ exists, $u \equiv 0$ in $\mathbb{R}^n\setminus\{0\}$ and $\int_{\mathbb{R}^n}u=1$? This leads to a contradiction because $\int_{\mathbb{R}^n}u = \int_{\mathbb{R}^n\setminus\{0\}}u$, so no such $u$ exists.
 A: Intuitively, because there a bump functions $\varphi$ with $\varphi(0) = 1$ but which are zero except on an arbitrary small interval $[-\epsilon,\epsilon]$ around $0$. That, however, is incompatible with the requirement that for some fixed $u$, $$
  \int_{\mathbb{R}^n} u(x)\varphi(x) \,dx = \varphi(0) = 1
$$
essentially because you can make that integral arbitrarily small by picking a suitably narrow $\varphi$.
For a formal proof, let $(\varphi_n)_{n\in\mathbb{N}}$ be a sequence of bump functions with $\varphi_n(0) = 1$, $\|\varphi_n\|_\infty = 1$ and $\textrm{supp}\, \varphi_n = [-\frac{1}{n},\frac{1}{n}]^k$ (i.e., $\varphi_n$ is zero outside the cube $[-\frac{1}{n},\frac{1}{n}]^k$).
You then on the one hand get $$
  \lim_{n\to\infty} \int_{\mathbb{R}^k} u \varphi_n 
  = \lim_{n\to\infty} \langle I_u, \varphi_n \rangle
  = \lim_{n\to\infty} \varphi(0) = 1 \text{.}
$$
But on the other hand, since obviously $u \varphi_n \to 0 $ pointwise as $n \to \infty$, you get by dominated convergence (note that $|u(x)\varphi_n(x)| \leq |u(x)|\cdot\|\varphi_n\|_\infty \leq |u(x)|$ for all $x \in [-1,1]^k$ and that $|u|\in L^1_{[-1,1]^k} \subset L^1_{\textrm{loc}}$) that $$
   \lim_{n\to\infty} \int_{\mathbb{R}^k} u \varphi_n
   = \lim_{n\to\infty} \int_{[-1,1]^k} u \varphi_n
   = 0 \text{.}
$$
This obviously is a contradiction.
A: Proof by @fgp is very nice. Here it's alternative proof:
Suppose that exist function $u(x) \in L^1_{\text{loc}}(\mathbb{R}^n)$ such that 
$$\int_{\mathbb{R}^n} u(x)\varphi(x)\, dx = \varphi(0), \quad \text{for all } \varphi \in \mathcal{D}(\mathbb{R}^n).$$
If $\varphi \in \mathcal{D}(\mathbb{R}^n)$ then also $x_1 \varphi(x)=x_1 \varphi(x_1,x_2,\ldots,x_n)\in \mathcal{D}(\mathbb{R}^n)$, so then $$\int_{\mathbb{R}^n} u(x)x_1 \varphi(x) \, dx =x_1 \varphi(x){\Large{|}}_{x=0}=0, \quad \text{for all } \varphi \in \mathcal{D}(\mathbb{R}^n).$$ 
According to du Bois-Reymond lemma that implies $x_1 u(x)=0$ almost everywhere, and then $u(x)=0$ almost everywhere, so we get 
$$\int_{\mathbb{R}^n} u(x)\varphi(x)\, dx = 0,$$
which is in contradiction with our initial assumption.
