Understand $\delta $ function, why do me write $\int \delta (x)dx$ instead of $\int d\delta $ ? Since $\int \delta (x)dx$ should be $0$ I have difficulties to understand delta function. We defined $\delta $ function as $$\delta (x)=\begin{cases}\infty&x=0\\ 0&\text{otherwise}\end{cases},$$
and $$\int_{\mathbb R}\delta (x)dx=1.$$
1) How can it be possible to have such definition since $\delta (x)=0$ a.e., and thus, the integral above should be $0$ ?
2) If we consider $\delta $ as a measure, then $\int_A f(x)d\delta(x) $ make completely sense for any set $A$. Now, I often see $$\int_A f(x)\delta (x)dx$$
for $$\int_A f(x)d\delta .$$
What ? This is not correct, is it ? Because $\int_A f(x)\delta (x)dx=0$ since $f(x-\delta (x)=0$ a.e. So why do we use this notation of $\int_A f(x)\delta (x)dx$ for $\int_{\mathbb R} f(x)d\delta (x)$ ? I'm really in truble with that. 
 A: It is indeed as @GEdgar mentioned. But, one way, to answer your first question and to give a better idea about the delta function we consider $\delta$ to be defined as follow,
$$\delta(x)=\lim\limits_{\varepsilon\rightarrow 0}\delta_\varepsilon(x),\quad \delta_\varepsilon(x)=\left\{
\begin{array}{lc}
0 & x<-\frac{\varepsilon}{2} \\
\frac{1}{\varepsilon} & -\frac{\varepsilon}{2}<x<\frac{\varepsilon}{2}\\
0&\frac{\varepsilon}{2}<x
\end{array}
\right.$$
then we take the integral of $\delta_\varepsilon$ over $\mathbb{R}$,
$$\forall\varepsilon>0,\quad \int_\mathbb{R}\delta_\varepsilon(x)dx=\int_{-\frac{\varepsilon}{2}}^{\frac{\varepsilon}{2}} \frac{1}{\varepsilon}dx=\frac{1}{\varepsilon}\bigg(\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\bigg)=1$$
Now we apply the limit when $\varepsilon\rightarrow 0$ and by applying the Dominated Convergence Theorem we get,
$$ 1=\lim\limits_{\varepsilon\rightarrow 0}\int_\mathbb{R}\delta_\varepsilon(x)dx=\int_\mathbb{R}\lim\limits_{\varepsilon\rightarrow 0}\delta_\varepsilon(x)dx=\int_\mathbb{R}\delta(x)dx$$
It is one way to understand the first mentioned property delta function, but of course the "Distribution Theory" gives a more rigorous and well-suited definition of the delta distribution and its properties.
A: why do me write $\int \delta (x)dx$ instead of $\int d\delta $ ?
Because we are physicists and not mathematicians!  
You say you are a mathematician?  Then you may write it correctly if you like.  I agree that this notation is harmful to beginners and non-mathematicians.  (Just see all the confused questions on it in math.se.)  But usage among physicists is unlikely to change.
