finite integrals where the integrand can approach infinity

Question

I asked this a while ago and then went onto look into measure theory and see whatever I can understand. Apologies as I have a poor background in mathematics.

So, my original query was when I was looking at the density function of the Gamma distribution as I show in this plot:

So here, the function blows up at $\theta = 0$ but it still integrates to 1. I can understand that around $\theta=0$, the function rises quickly and then it has to fall down to compensate for it.

Now, I am trying to understand the contribution of the $\theta=0$ point to the integral. From what I can understand, the contribution of $\theta=0$ is actually 0 to the integral. This is because the set $\theta=0$ is a closed set and has measure 0 because it only has one element i.e. it is finite and countable. Is this understanding correct or am I to think along the lines that the function value grows out of bounds but it is multiplied by $dx$ at 0 and hence this contribution is somehow very small but non-zero.

Another question I have is if the function was modified so that as $\theta$ approaches 1, the function again blows up. Would the integral still be finite? My reasoning is that the points at which the function goes to infinity forms the set $\{0, 1\}$ which is still finite and countable. So the contribution from those points is exactly 0. This reasoning somehow also makes sense to me as the probability for a particular point is exactly 0.

Answer 1

First question: yes. A point always contributes zero to a Lebesgue integral. The point itself is not responsible for what the function does around that point.

Second question: it depends on how fast exactly it blows up. But if it is the same kind of singularity as in $0$, then the integral is finite.

The functions $\frac1{x^k}$ where $k$ is an integer, $k\geq 1$, blow up too slowly at $0$ for the integral to be finite. Whereas the functions $\frac1{x^{1/k}}$ where $k$ is an integer, $k> 1$, blow up more abruptly at $0$ and the integral is locally finite. You can transpose these examples to $x=1$ by composing these functions with $x\mapsto x-1$.