Is this interpretation of the Dirac-measure property correct? First and foremost, apologies in advance for using an abuse of notation by placing the Dirac measure inside an integral for which I was told that this should not be done from a previous question asked by me. But given the circumstances, I have no choice. 
This is essentially a word by word copy of an interpretation given on page 1 of these Berkeley notes: 


The important property of the delta function is the following relation
  $$\displaystyle\int f(t) \delta(t) \, \mathrm{d}t = f(0)$$  for any
  function $f(t)$. This is easy to see. First of all, $\delta(t)$
  vanishes everywhere except when $t = 0$. Therefore, it does not matter
  what values the function $f(t)$ takes except at $t = 0$. You can then
  say $f(t)\delta(t) = f(0)\delta(t)$. Then $f(0)$ can be pulled outside
  the integral because it does not depend on $t$, and you obtain the
  r.h.s.


Here's the problem, it was my understanding that $$\delta(t) = \begin{cases} 0 & \space \mathrm{for} \space t \ne 0 \\\infty&\ \mathrm{for} \space t = 0 \end{cases} $$ So by my logic this means that $\delta(0)=\infty$ and therefore undefined; which implies that when $t=0$ $$\displaystyle\int f(0) \delta(0) \, \mathrm{d}t = \displaystyle\int f(0) (\infty) \, \mathrm{d}t$$ which is manifestly not true and certainly not equal to $f(0)$.  
Clearly I am missing the point of this argument, so if someone would be kind enough to explain it to me I would be most grateful.
Thank you in advance.
Regards, 
Blaze
 A: The first time I encountered the Dirac delta was when I did my first (and only) course in electrostatics. It was introduced in the following way 
$$
\delta(x) = \begin{cases} 0 & \space \mathrm{for} \space x \ne 0 \\\infty&\ \mathrm{for} \space x = 0 \end{cases}
$$
and $$
\int_{-\infty}^{\infty} \delta(x) dx = 1
$$
We were told that we should think of this as an enormous high spike at $x=0$ which one might construct by taking the limit of rectangles with height $n$ and width $1/n$. The interesting property of the Dirac delta is the following
$$
f(x) \delta(x) = f(0)\delta(x)
$$
for all "ordinary functions" i.e. not another delta function (sometimes limited to continuous functions). This should be clear since $f(x)\delta(x)$ has to be $0$ when $x \neq 0$. Now we can use this fact on the integral
$$
\int_{-\infty}^{\infty} f(x)\delta(x) dx = \int_{-\infty}^{\infty} f(0) \delta(x) dx 
$$
and since $f(0)$ is a constant
$$
\int_{-\infty}^{\infty} f(0) \delta(x) dx  = f(0) \int_{-\infty}^{\infty} \delta(x) dx = f(0) 
$$
I hope this makes some intuitive sense to you.
Now I don't like this explanation since I've been given a more rigourous one in a recent course on measure theory and the Lebesgue integral. Here the Dirac delta can be defined as a measure on a set in a much nicer way that maintains this property when integrating.
