 # Let ( <mi mathvariant="normal">&#x03A9;<!-- Ω --> , <mrow class="MJX-TeXAtom-ORD"> Kyle Sutton 2022-07-10 Answered
Let $\left(\mathrm{\Omega },\mathcal{F},\mathbb{P}\right)$ be a probability space and $X:\left(\mathrm{\Omega },\mathcal{F}\right)\to \left(\mathbb{R},{\mathcal{F}}_{E}\right)$ a random variable that admits a density function ${f}_{X}:\mathbb{R}\to \mathbb{R}$.
I have some questions about the information given by the cumulative distribution function of X (${F}_{X}:\mathbb{R}\to \left[0,1\right]$).
i) Does the CDF determine ${f}_{X}$ uniquely ?
ii) Does the CDF determine the law of X uniquely ?
iii) Can we find a subset of events that uniquely determine the CDF ?
For i), I would say that as ${F}_{X}\left(t\right)={\int }_{-\mathrm{\infty }}^{t}{f}_{X}\left(x\right)dx$, the density function is the derivative of the CDF and therefore the CDF can only define one density function. But I feel like maybe we could find a counterexample by considering Lebesgue measure and taking sets of the form (a,b) and [a,b].
For ii), as $\mathrm{\forall }x\in \mathbb{R}$, $\mathbb{P}\left(X=x\right)={F}_{X}\left(x\right)-{F}_{X}\left(x-\right)$ and $\mathbb{P}\left(X\le x\right)={F}_{X}\left(x\right)$, I am tempted to say that the law of X is indeed by definition determined uniquely by the CDF.
For iii), maybe we could take all events of the form $\left\{X\in \left(a,b\right):a,b\in \mathbb{R}\right\}$ but I'm not quite sure about that, and particularly regarding the uniqueness.
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By subtraction, the CDF $F\left(x\right):=P\left(X\le x\right)$ determines the probabilities $P\left(X\in \left[a,b\right]\right)$ for every $a,b\in \mathbb{R}$. By an approximation theorem, such as Caratheodory's theorem, this information determines $P\left(X\in E\right)$ for every $E\in B\left(\mathbb{R}\right)$. So (ii) is true.
(i) follows (though, as always, the density is unique only up to almost everywhere equality).

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