Value of a generalised probabilty density function with a dirac delta function at a single point I thought I understood the dirac delta function until I came across a question where I needed to evaluate the probability density function at a single point.
Probability density function:
$$f_X(x) = \frac{1}{2}\delta(x-1) + \frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$
I would like to find the value of the function at $x=1$.
I know that the dirac delta function is used to account for the discontinuities in the cumulative distribution function, and that the dirac delta function is defined as:
$$\delta(x)  = \begin{cases}
    \infty,&  x = 0\\
    0,& otherwise.\end{cases}
$$
I could choose to ignore the discontinuity at $x=1$, but that could result in
$$f_X(x=1) = \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}},$$
$$f_X(x=1) = \frac{1}{2} + \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}}$$
or any value in between.
But, according to the definition of the dirac delta function, at point $x=1$ the function could also evaluate to
$$f_X(x=1) = \infty + \frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}} = \infty$$
This could be reasonable because the dirac delta function is not a traditional function, therefore I could accept a definition, if needed. The problem arises when I try to calculate conditional probabilities in any intervals containing $x=1$. For e.g., if I accept that $f_X(x=1) = \infty$, then
$$P(X=1|X \geq 1) = \frac{P(X=1)}{\int_{1}^{+\infty}(\frac{1}{2}\delta(x-1) + \frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{2}})} = \frac{P(X=1)}{\frac{1}{2\sqrt{2\pi}} \int_{-\infty}^{1} e^{-\frac{x^2}{2}}} = \infty,$$
which is clearly nonsensical. So then the questions arise, what is the value of $f_X(x=1)$ and why?
 A: The Dirac delta is not a function.  It is not defined by its values.  Otherwise there would be no difference between $\delta (x) $ and $5 \delta (x) $.  The Dirac delta is a generalized distribution and does not have values at all.  The definition you give for it is incorrect; there is a way to twist and interpret it into a true statement, but it is very misleading at best.  A generalized distribution is defined by its integrals.  The probability distribution you gave is not a function and does not have a value at 1 at all.  Because of the $\delta$, the entire expression is a generalized distribution and not a function.
A: 
I would like to find the value of the function at x=1.

Your $f(x)$ is a "mixed density", it is not exactly a density function because your random variable is not absolute continuous. It has a positive probability mass in $X=1$ so you have
$$\mathbb{P}[X=1]=\frac{1}{2}$$
For the rest of the support you have a Standard Gaussian density thus, for example:
$$\mathbb{P}[X\geq 1]=\frac{1}{2}+\frac{1}{2}\left[1-\Phi(1)\right]\approx0.58$$
So for your question you will have
$$\mathbb{P}[X=1|X\geq 1]\approx\frac{50}{58}$$

First observation

Here is the correct (better, here is what your text means as) definition of your Dirac Delta
$$\delta(x-1)=\begin{cases}
1,  & \text{if $x=1$ } \\
0, & \text{elsewhere}
\end{cases}$$
For detail, read here in the paragraph "Generalizations"

Second observation

In these cases it is better to define the random variable with its CDF that always exists rather then a "density" that will not exist.
Anyway I know that in some engineering applications, say telecommunication or Signal's Analysis they often use mixed density with impulse (Dirac delta), Rectangular (uniform) response et cetera et cetera
