Why is probability of any outcome on an uniform distribution $[0,1]$ zero?

Question

Consider the uniform distribution on $[0,1]$. Let $s\in [0,1]$ be any outcome. What is $P\{s\}$ ?

How I understand it:

We know that $P([a,b]) = b -a \quad$ where $0 \leq a \leq b \leq 1$

And $s$ is any outcome, so here $a$ and $b$ can be any numbers between zero and one.

So why if we pick any outcome $s$ we get $P(\{s\}) = 0$

Does anybody know how to explain this ?

Pick your favorite number $s$ in $[0,1]$, say $s=1/\pi$. The probability to get an outcome equal to $s$ up to the $i$th decimal digit is roughly $10^{-i}$. If you want the outcome to be exactly $s$ however you need the intersection of all these events, which has measure zero. I am sure there are plenty of similar answers (also on MSE). — Giulio R, Sep 29 '21 at 11:17
How would you explain it using the formulas you have written? What values of $a$ and $b$ give you $P({s})$? — Joe, Sep 29 '21 at 11:20
$P(X=s)=P(X\le s)-P(X\lt s) = (s-a)-(s-a)=0$ for $s \in [0,1]$ — Henry, Sep 29 '21 at 12:01
Does this answer your question? Why is the probability that a continuous random variable takes a specific value zero? — Peter O., Sep 29 '21 at 13:15

score 2 · Accepted Answer · answered Sep 29 '21 at 11:22

2

Choosing $a,b$ such that $s\in[a,b]\subseteq[0,1]$ we have:$$P(\{s\})\leq b-a$$

Here we can make $b-a$ as small as we want, so assumption $P(\{s\})>0$ leads to a contradiction.

answered Sep 29 '21 at 11:22

drhab

151,093

score 2 · Answer 2 · answered Sep 29 '21 at 11:33

2

Notice in the definition $0 \leq a \leq b \leq 1$.
Do you see $\le$ in there between $a$ and $b$?
So, given any $s \in [0,1]$, take $a=s, b=s$ and conclude $$ P\big(\{s\}\big) = P\big([s,s]\big) = s-s = 0 . $$

answered Sep 29 '21 at 11:33

GEdgar

111,679

So basically we are looking at the intersection not the union right ? – VLC Sep 29 '21 at 11:36
1

I see neither intersection nor union. We apply one instance of the given formula. – GEdgar Sep 29 '21 at 11:40

Why is probability of any outcome on an uniform distribution $[0,1]$ zero?

2 Answers2