Why is a probability density function nonnegative? Let $X$ be a random variable and its density $f$ be defined to be the derivative of its distribution function $F$, i.e. $$\Pr(a< X\le b)=F(b)-F(a)=\int_a^bf(x)\operatorname{dx}$$ Now let $$N:=\left\{x:f(x)<0\right\}$$ Of course, we must have $$0\le \Pr(X\in N)=\int_Nf(x)\operatorname{dx}\le 0$$ which forces $X\notin N$, almost surely. But I don't see why we need $N=\emptyset$ in order for $f$ to be a density function.
 A: $f(x)$ taking negative values would mean that $F(x)$ wouldn't be monotonically increasing, thus allowing for absurd scenarios like $\Pr(X \in I_1) \ge \Pr(X \in I_2)$ despite $I_1 \subset I_2$. Or in general: A logical disjunction resulting in reduced probability, i.e. $\Pr(A \cup B) \le \Pr(A)$, which contradicts everything probability theory stands for.
A: $N$ is where your density function is negative.  If $Pr(X \in N) > 0$, then it would not be the case that $Pr(X\in N) = \int_N f(x)dx$, since the left-hand-side would be positive, and the right-hand-side negative. But part of what we want the density function to do is for any set $A$ that can be assigned a probability, for it to be $Pr(X \in A) = \int_A f(x)dx$. 
If we allow the density function to be negative (i.e. $Pr(X \in N) > 0)$, your example shows how this breaks.
A: Be careful! You swaped definitions of a density and a distribution function. You stated that a density is a primitive of its distribution function but it is otherwise. So, a density is a derivative of its distribution function.
Hence, value of a density in each point gives a slope of its distribution function and every distribution function is non-decreasing in each point (obviously it cannot be true that $F(b)=P(X<b)>F(a)=P(X<a)$ when $a>b$).
Thus, derivation of every distribution function (which is its density) must be non-negative (which means $N=\emptyset$ in your notation).
