Dirac Delta function at a point From my understanding of the Dirac Delta function, it is infinitely thin and has a value of infinity at only a particular point. I also learned that
$$\int_{-\infty}^{\infty} \delta(x-a) dx = 1$$
What if we have a case where we are faced with the integral:
$$\int_{-\infty}^{a} \delta(x-a) dx$$
How does the fact that the function is infinitely thin at a affect the result?
 A: Coming from a measure-theoretic point of view, one can write the following: 
$\int_{-\infty}^{+\infty}\delta(x-a)dx = \int_{\mathbb R} 1 d \mu_a(x)$, where $\mu_a$ denotes the point measure with mass $1$ at $a$ and $0$ everywhere else. 
Now, if we want to integrate over an interval $I$, we integrate the characteristic function $\chi_I(x)$ over $\mathbb R$:
$$\int_I \delta(x-a)dx= \int_{\mathbb R}\chi_I(x)d\mu_a(x).$$
And here comes a important distinction which we dont need to make when using the Riemann integral: If $a$ is a border point of the interval $I$, do we consider the open interval $I_1 =(-\infty,a)$, or the interval $I_2=(-\infty,a]$ ? Because the mass of $\mu_a$ lies at $a$, this is an important distinction:
$$\int_{(-\infty,a)} \delta(x-a)dx= \int_{\mathbb R}\chi_{(-\infty,a)}(x)d\mu_a(x) = 0 $$
while 
$$\int_{(-\infty,a]} \delta(x-a)dx= \int_{\mathbb R}\chi_{(-\infty,a]}(x)d\mu_a(x) = 1.$$
So we have to be very careful with what we mean by $\int_{-\infty}^a\delta(x-a)dx.$
Summing up: The fact that it's 'infinitely thin' translates to the measure having a point mass, and this means that the inclusion or the exclusion of single points will change the value of the integral. 
A: The Dirac Delta function is a symmetrical function.Thus $$\int_{-\infty}^{a} \delta(x-a) dx = \int_{a}^{+\infty} \delta(x-a) dx = \frac{1}{2}$$  Furthermore $$\int_{a^-}^{a^+}\delta(x-a) dx = 1$$ And $$\int_{a^-}^{a}\delta(x-a) dx = \int_{a}^{a^+}\delta(x-a) dx = \frac{1}{2}$$
