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I think it is not too dumb a question..

I get a definition of uniform distribution from wikipedia. It says that uniform distribution is defined such that density is 1 at the interval $[0,1]$ and zero everywhere. Well, if we integrate them throughout all real line, Indeed we get the integration 1. However, if we consider another density which is 1 at the interval $(0,1)$ and zero everywhere. We still get the integration 1. It seems it is also a valid probability density.. My question is that is this new density also uniform distribution?

Maybe more general, how about density with value 1 in $(0,1]$ or in $[0,1)$.

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    $\begingroup$ Try checking the definition carefully. $\endgroup$
    – sam wolfe
    Oct 11, 2017 at 23:32

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Let $\mathsf{P}$ be a probability measure on $\mathbb{R}$ s.t. $\mathsf{P}(B)=m(B\cap[0,1])$ for a Borel set $B$, where $m$ is the Lebesgue measure on $\mathbb{R}$ ($\mathsf{P}$ corresponds to the uniform prob. distribution on $[0,1]$). Since $\mathsf{P}\ll m$, there exists the Radon-Nikodym derivative $f=d\mathsf{P}/dm$, which is unique up to $m$-null sets (in this case $f=1_{[0,1]}$ a.e.). It means that any $g$ s.t. $m(\{g\ne 1_{[0,1]}\})=0$ is a version of the density of $\mathsf{P}$ w.r.t. $m$ (for example, we may take $g=1_{[0,1]\setminus \mathbb{Q}}$).

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  • $\begingroup$ Thank you, Indeed help me a lot. One more question, Is the probability measure and the corresponding radon-Nikodym derivative which is unique up to $m$-null sets a one-to-one mapping? $\endgroup$
    – Ethanabc
    Oct 12, 2017 at 13:31
  • $\begingroup$ $f$ is a unique function which gives $\mathsf{P}(B)=\int_B f dm$ for all Borel $B$. $\endgroup$
    – user140541
    Oct 12, 2017 at 16:14
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The uniform distribution is a continuous one, thus the probability of any particular value is $0$. It does not matter if the bounds are strict or not, it is still the same distribution.

$$P(X=x)=0, \forall x$$ so it does not matter if you include $P(X=0)$ or $P(X=1)$ as these are both $0$

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  • $\begingroup$ Actually you also have the discrete uniform distribution. $\endgroup$
    – sam wolfe
    Oct 11, 2017 at 23:32
  • $\begingroup$ I am aware of that, but the discrete uniform would not make sense given what he has said in the description of it with $0<x<1$ as this has no integer values in between. $\endgroup$
    – Harry Alli
    Oct 11, 2017 at 23:34
  • $\begingroup$ Ah, that's true! $\endgroup$
    – sam wolfe
    Oct 11, 2017 at 23:36
  • $\begingroup$ @Harry Alli, Do you mean that probability distribution is uniquely determined by the distribution function (in one dimension, it is the CDF function). It is possible that different densities will give us the same probability distribution.. $\endgroup$
    – Ethanabc
    Oct 11, 2017 at 23:36
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It's the same distribution either way, since its integral over any set is the same.

But since you mention "support", we should look at the definition of that term. A number $a$ is a member of the support of a distribution if that distribution assigns positive probability to every interval $(a-\varepsilon,a+\varepsilon),$ no matter how small $\varepsilon>0$ is.

That implies $0$ and $1$ are both members of the support of this distribution, so the support is $[0,1].$ The support is always a closed set.

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