Dirac Delta Function "removing integral" I am trying to proof the following  identity:
\begin{align}
 \delta(ax) = \frac{1}{|a|}\delta(x).
\end{align}
where $\delta$ represents the dirac delta function.
I found that
\begin{align}
  \int_{-\infty}^\infty  f(x) \delta(ax) \,\,dx = \frac{1}{|a|} \int_{-\infty}^\infty f(x)\delta(x)\,\,dx.
\end{align}
Consequently we get
\begin{align}
 \delta(ax) = \frac{1}{|a|}\delta(x).
\end{align}
by "removing" the integrals. Is this step actually valid?
 A: As pointed out by @GEdgar in the comments, equality here is in the sense of distribution, and integration of a Dirac delta against a function is not really a Riemann or Lebesgue integral. It's an abuse of notation. But what does all of that mean?
The space of distributions, often denoted $\mathcal{D}^{'}(\mathbb{R})$, is the space of all maps $f: \mathcal{C}_c ^\infty (\mathbb{R}) \to \mathbb{R}$ that are linear and continuous (whatever that means for now) where $\mathcal{C}_c ^\infty (\mathbb{R})$ is the space of smooth functions with compact support. For example, the dirac delta takes any function $\phi \in \mathcal{C}_c ^\infty (\mathbb{R})$ and maps it to its value at zero: $\phi(0) \in \mathbb{R}$. The action of dirac delta on $\phi$ is denoted: $\langle \delta, \phi \rangle$. It can be shown that such a map is linear and continuous, and hence, the dirac delta is a distribution. If a map $f$ is smooth enough, for example, a continuous function, then you can write: $\langle f, \phi \rangle = \int_{-\infty} ^{\infty} f \phi \; dx$ for any $\phi \in \mathcal{C^\infty _c(\mathbb{R})}$. However, this is not necessarily the case. With that in mind, let's move on to clarify your question.
Two distributions $f,g$ are said to be equal if $\forall \phi \in \mathcal{C^\infty _c(\mathbb{R})}$, we have:
$$ \langle f, \phi \rangle = \langle g, \phi \rangle
$$
In other words, both distributions induce the same effect on $\phi$. So, the equality: $\delta(ax) = \frac{1}{|a|} \delta(x)$ actually reads: the action of the distribution $\delta(ax)$ on an arbitrary fixed $\phi$ is equal to the action of the distribution $\frac{1}{|a|} \delta(x)$ on $\phi$. Formally; $\langle \delta(ax), \phi(x) \rangle = \langle \frac{1}{|a|}\delta(x), \phi(x) \rangle $. In order to formally prove this result we need to know what is the composition of a distribution.
The left hand-side of your equality is $\delta(ax)=\delta \circ g$ where $g(x)=ax$. Now that we know what a distribution is, we can move on and ask ourselves what the composition of the Dirac delta distribution with a function $g$ is. Is the composition of a distribution with a function $g$ a distribution? It turns out that the answer is yes if $g$ and its inverse are smooth. Which is true for our case (assuming $a \neq 0)$. This composition is defined as:
$$\begin{aligned}
\langle \delta \circ g, \phi \rangle &= \langle \delta, \phi \circ g^{-1} |\text{det } dg^{-1}| \rangle \\
& = \langle \delta, \phi (\frac{y}{a}) \frac{1}{|a|} \rangle \\
& = \frac{1}{|a|} \langle \delta, \phi (\frac{y}{a})  \rangle \\
& = \frac{1}{|a|} \phi(0) \\
& = \langle \frac{1}{|a|}\delta, \phi \rangle \\
\end{aligned}
$$
If we allow for an abuse of notation, the definition of a composition of a distribution with a smooth function is equivalent to changing the variabe of integration!
$$
\int_{-\infty}^{\infty} \delta(g(x)) \phi (x) \; dx = \int_{-\infty}^{\infty} \delta(y) \phi (\frac{y}{a}) \frac{1}{|a|}  \; dy
$$
A: Shortly: yes, but talking about "removing the integrals" is an oversimplification.
A bit more complete answer: $\delta(x)$ is defined by
$$\int_{-\infty}^\infty f(x)\,\delta(x)\,dx=f(0)$$
(it must hold for any sufficiently regular $f(x)$).
You found that, for any sufficiently regular $f(x)$,
$$\int_{-\infty}^\infty f(x)\,\delta(ax)\,dx
 = \frac{1}{|a|}\int_{-\infty}^\infty f(x)\,\delta(x)\,dx
 = \frac{1}{|a|}f(0)\;,$$
so, after multiplying both sides by $|a|$, you obtain
$$\int_{-\infty}^\infty f(x)\,|a|\delta(ax)\,dx=f(0)\;.$$
Since it holds for any $f(x)$, it means (by definition) that
$$|a|\delta(ax)=\delta(x)\;.$$
