Proving the sifting property of the Dirac delta

Question

How to prove the following property of the Dirac delta?

$$ f(x)= \int_0^1 f(a) \delta (x-a)da $$

for $ 0 < x < 1 $

Answer 1

In general, the composition of a distribution with a nice function is defined so that the change of variable formula holds. In particular, the distribution $\delta(x-a)$ verifies by definition $$ \int_{-\infty}^\infty \varphi(a)\,\delta(x-a)\,\mathrm d a = \int_{-\infty}^\infty \varphi(x-y)\,\delta(y)\,\mathrm d y = \varphi(x-0) = \varphi(x) $$ for any function $\varphi$ continuous at $x$. In your case, to take care of the boundaries, you can use the above property with $\varphi = \mathbb{1}_{[0,1]}\, f$ where $\mathbb{1}_{[0,1]}$ is the indicator function of the set $[0,1]$ and get $$ \int_{0}^1 f(a)\,\delta(x-a)\,\mathrm d a = \int_{-\infty}^\infty \mathbb{1}_{[0,1]}(a)\, f(a)\,\delta(x-a)\,\mathrm d a = \mathbb{1}_{[0,1]}(x)\, f(x) $$ if $x≠0$ and $x≠1$. In particular, if $x\in(0,1)$ $$ \int_{0}^1 f(a)\,\delta(x-a)\,\mathrm d a = f(x). $$