Integral of Probability Density Function over the Reals is Equal to One

Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $X$ be an absolutely continuous random variable.

Let $f_X$ be a probability density function for $X$.

Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$.

Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$.

Then:


 * $\ds \int_\R f_X \rd \lambda = 1$

where $\ds \int_\R f_X \rd \lambda$ denotes the Lebesgue integral of $f_X$.

Proof
Let $P_X$ be the probability distribution of $X$.

From the definition of a probability density function, $f_X$ is a Radon-Nikodym derivative of $P_X$ with respect to $\lambda$.

We then have: