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Let $(\Omega,\mathcal F,P)$ be a probability space. Let $\xi:\Omega\to\mathbb R$ be a random variable. Let $P_\xi$ be the distribution of $\xi$. Suppose $P_\xi$ is a continuous distribution. (i. e. There is a Borel function $f_\xi:\mathbb R\to\mathbb R$ such that for all $B\in\mathcal B(\mathbb R)$, $P_\xi(B)=\int_Bf_\xi dx$. ($f_\xi$ is the density of $\xi$.)) Is $f_\xi$ nonnegative almost everywhere? (i. e. Is there a Lebesgue measurable set $A$ such that $\mu(A)=0$ and $f_\xi(x)\geq0$ for all $x\in\mathbb R-A$, where $\mu$ is the Lebesgue measure?)

Here's my attempt:

Let $C=f_\xi^{-1}((-\infty,0))$. It is clear that $C$ is a Borel set. So if we prove that $\mu(C)=0$, then we are done.

Assume $\mu(C)>0$.
Let's prove that $\int_C f_\xi dx<0$ so that $P_\xi(C)<0$, a contradiction. Assume $\int_C f_\xi dx\geq0$. By the definition of $C$, it is clear that $\int_C f_\xi dx$ can't be greater than $0$, so we have $\int_C f_\xi dx=0$. Let $f_\xi^-:=-\operatorname {min}\{f_\xi,0\}$. Choose a sequence $(s_n)$ of measurable, nonnegative simple functions which monotonically increases and $s_n\to f_\xi^-$ pointwise. By the Lebesgue's monotone convergence theorem, we have $\int_Cs_ndx\to\int_Cf_\xi^- dx$. So $\int_Cs_ndx$ should be $0$ for all $n$. I tried to derive a contradiction here using the fact that $f_\xi^-(x)>0$ for all $x\in C$, but I couldn't do it.

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What follows after "Assume $\mu(C)>0$" in your question should be followed by this:

$C=\bigcup_{n=1}^{\infty}C_n$ where $C_n:=f_{\xi}^{-1}((-\infty,-\frac1n])$ so that $C_1\subseteq C_2\subseteq\cdots$ and consequently $\mu(C_n)\uparrow\mu(C)$

Then assuming $\mu(C)>0$ leads to $\mu(C_n)>0$ for $n$ large enough and consequently: $$P_{\xi}(C_n)=\int_{C_n} f_{\xi}\;d\mu\leq -\frac1n\mu(C_n)<0$$A contradiction, and you are ready.

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You are making things complicated at the end. You have proved that $\int_C f_{\xi} (x) dx=0$. But actually, if you look at the proof of this you will get the same conclusion if $C$ is replaced by any Borel subset of $f_{\xi}^{-1} (-\infty,0)$. Now take $C=f_{\xi}^{-1} (-\infty,0) \cap \{x:f_{\xi} (x) <-\frac 1 n\}$ and conclude that this $C$ has measure $0$ for each $n$. Take union over $n$ to complete the proof.

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