Lebesgue measure on $\mathbb{R}$ is not a probability measure So I'm not a math student hence the (probably rather simple) question in the title. I could not find an answer in the slides or the internet. 
My intuition is that the elements (ie. reals) of  $\mathbb{R}$ are not countable, hence no probability measure can be established. Am I correct and is this a sufficient explanation? 
 A: Probability measures give the entire space measure $1$. This has nothing to do with countability (although the existence of a uniform probability measure does have something to do with countability).
Since the Lebesgue measure of an interval is equal its length, it is fairly easy to see that there are sets of measure greater than $1$ and that the entire space has infinite measure. Therefore this is not a probability measure.
A: The total probability must always be $1$, so any probability measure must give this value as the measure of the whole space.  Obviously this fails for the Lebesgue measure, since the volume of $\mathbb{R}$ is infinite.
But this has nothing to do with the uncountability of $\mathbb{R}$.  For example, we could weight the Lebesgue measure by a normalised Gaussian; this gives a probability measure on $\mathbb{R}$.
A: No, the whole point of basing probability theory on measure theory is to overcome precsily those problems, and to clearly define probabilities for subsets of uncoutable sets. What you're looking for is much simpler.
For a probability measure, you want that the measure (i.e. the probability!) of the whole set is $1$. In other words, for a probability measure $\mu$ on $\mathbb{R}$, you want that $\mu(\mathbb{R})=1$. The lebesgue measure $\lambda$, however, has $\lambda(\mathbb{R}) = \infty$.
You want $\mu(\mathbb{R})=1$ for a probability measure on $\mathbb{R}$ simply because, as it's name implies, a probability measure measures probabilities of an outcome being part of a subsets (called events in probability theory). The probability of $x > 0$ would e.g. be $\mu(\{x:x > 0\})$. Since $x \in \mathbb{R}$ imposes no restriction on $x$ if the $\mathbb{R}$ is the whole probability space, $\mu(\mathbb{R})$ has to be $1$, since $x$ will be an element of $\mathbb{R}$ with probability $1$.
