# Sufficiency for being a probability density function

Once my professor told us in passing that a non-negative integrable (Riemann or Lebesgue) function that integrates to one over its support need not be a probability density function. I have since tried to find counterexamples where this is true, but have failed. Is there any such counterexample? Also, is there a theorem establishing sufficient conditions on a function to be a pdf?

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Yes: if the integral of the function is not $1$. – Did Mar 20 '12 at 9:08
Delta functions? – yohBS Mar 20 '12 at 9:10
@DidierPiau: Oops, sorry! Adding that clause as well... – Bravo Mar 20 '12 at 9:14
Then the passing remark as you recall it is wrong. Maybe the remark was in fact that not every PDF corresponds to such a function. – Did Mar 20 '12 at 9:57

If $f:\mathbb{R}\to \mathbb{R}$ is a Borel function that satisfies $f(x)\geq 0$ for all $x\in\mathbb{R}$ and $\int_\mathbb{R} f d\lambda =1$, then $$P(A)=\int_A f d\lambda,\quad A\in \mathcal{B}(\mathbb{R})$$ defines a probability meausure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Now let $(\Omega,\mathcal{F},P)=(\mathbb{R},\mathcal{B}(\mathbb{R}),P)$ be your probability space and define a random variable $X:\Omega \to \mathbb{R}$ by the identity mapping, i.e. $X(\omega)=\omega, \;\omega\in\Omega$. Then the distribution of $X$ is $P\circ X^{-1}=P$ which has density $f$.

So I guess you would have to look at non-measureable functions $f$ to find your counterexample.

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You mean a non-Borel-measurable function that is Lebesgue integrable and integrates to unity? – Bravo Mar 20 '12 at 10:15
Oh, so you mean, $X$ is a RV, $f_X(x)$ is its pdf, over some set $A$, $\int_A f d\mu$ exists, but $P(X \in A)$ does not exist. – Bravo Mar 20 '12 at 10:18
All I'm saying is that if $f$ is non-negative, integrable and integrates to one and furthermore is Borel-measureable, then it is a probability density function for some random variable $X$. So if you want to find a function $f$ that is non-negative, integrable and integrates to one such that $f$ is not a pdf for some random variable $X$ then it can not be Borel-measureable. But there are functions that are Riemann-integrable without being Borel-measureable, so maybe that's where you should find your counterexample. – Stefan Hansen Mar 20 '12 at 10:43