Is $\mu$ a unique probability measure? Problem
Let $\mu$ be a measure on $(\mathbb{R}, \mathscr{B}(\mathbb{R})),$ which fulfills
$$  \mu([x,\infty)) =
\begin{cases}
\frac{1}{x+1},  & x>0 \\
1, & x\leq0
\end{cases}$$
Is $\mu$ a unique probability measure on  $(\mathbb{R}, \mathscr{B}(\mathbb{R}))?$
I don't really have a qualified attempt, because i can't find any theory in our book, and there really isn't anything when i search for a similar example.
Also, what does $[x,\infty)$ in the argument for $\mu([x,\infty))$ mean?
Can anybody lead me in the right direction?
 A: Found a related thing: Show that $\mu((-\infty,x])$ is a unique measure

Part1 (intuition) - From $\mu$'s value at inputs of $[x,\infty)$, we can get $\mu$'s values at other inputs...hopefully every Borel set:
$[x,\infty)$ is a closed interval from $x$ until $\infty$. You know Gasai the output value assuming you input a closed interval $[x,\infty)$ into $\mu$. What if you input something that doesn't look like $[x,\infty)$? Use the properties of measure!

*

*If I input $[3,4)$, then I can express $[3,4)$ as $[3,4) = [3,\infty) \ \setminus [4,\infty)$. Then $\mu([3,4)) = \mu([3,\infty) \ \setminus [4,\infty)) = \mu([3,\infty)) - \mu([4,\infty)$, by additivity (regular additivity not countable additivity!) of measure.


*If I input $[3,4]$, then I can express $[3,4]$ as $[3,4] = \{4\} \cup [3,4) = \{4\} \cup ( [3,\infty) \ \setminus [4,\infty))$. Then $\mu([3,4]) = \mu([3,4)) + \mu(\{4\})$.
As for $\mu(\{4\})$ ugh... Hmmm...ahh use of continuity of measure. We have
$$\mu([4,4+\frac1n)) = \frac{\frac1n}{4(4+\frac1n)}$$
and
$$\{4\} = \bigcap_{n=1}^{\infty} [4,4+\frac1n)$$
and
$$[4,4+\frac1n) \supseteq [4,4+\frac1{n+1})$$
so continuity of measure gives us
$$\mu(\{4\}) = \lim \frac{\frac1n}{4(4+\frac1n)} = \frac0{16} = 0$$
Sooo we can similarly get $\mu$'s values from $(x,\infty)$, $(-\infty,x)$, $(-\infty,x]$, $(a,b)$ and $(a,b]$ soooo I guess we can get $\mu$'s value at any Borel set.
Part2 - Proving it's a probability measure given it's a measure:
Easy. More than 1 way to do this I believe. 1 way I think is continuity of measure
$$\mu(\mathbb R) = \mu(-\infty,\infty) $$
$$= \mu(\bigcup_{n=1}^{\infty} [-n,\infty))$$
$$= \lim_{n \to \infty} \mu( [-n,\infty))$$
$$= \lim_{n \to \infty} 1$$
$$= 1$$
Part3 - Proving it's unique given it's a probability measure:
Hmmm...I know there's a uniqueness lemma that says if measures agree on a $\pi$-system $\mathscr A$, then they agree on the $\sigma$-algebra $\sigma(\mathscr A)$.
From David Williams' Probability with Martingales:



I think what you have to show here is if you have some other probability measure $\nu$ with the same formula as $\mu$ on $[x,\infty)$ then it'll have the same formula as $\mu$ on all Borel sets... ugh...not so sure... I think once you've shown it is a probability measure then it is a unique probability measure at least the way I did above because you can get the value of $\mu$ at any Borel set hmmmmm...
Well... $\mathscr B(\mathbb R) = \sigma(\mathscr A)$ with $\mathscr A = \pi(\mathscr R) := \{(-\infty,a]\}_{a \in \mathbb R}$, so just get the formula for $\mu$ on $\pi(\mathscr R)$, i.e. find the value of $\mu((-\infty,x])$ for every $x$ and then...yeah by the uniqueness lemma this $\mu$ is unique.

*

*For $x \le 0$, $\mu((-\infty,x]) = 0$


*For $x > 0$,
$$\mu((-\infty,x]) = \mu((0,x]) = \mu([0,x)) = \mu([0,\infty)) - \mu([x,\infty)) = 1 - \frac1{1+x} = \frac{x}{1+x}$$
Edit:
Wait i think it's unique simply because... if you have another measure $\lambda$ that satisfies the same property then $\mu$ and $\lambda$ will agree on $\pi (\mathbb R)$ ? Idk. I think it should be obvious, but I'm not really sure how to say this.
