# Bounded function with finitely many discontinuities is integrable $\overset{?}{\Rightarrow}$ density of continuous distribution function is not unique

The density function of the distribution function of a continuous random variable is not uniquely defined.
A new density function can be obtained by changing the value of the function at finite number of points to some non-negative value, without changing the integral of the function. We then get a new density function for the same continuous distribution.

Does this follow from the theorem-

A bounded function with finite number of discontinuities over an interval is Riemann integrable.

or is there a different theorem supporting the above claim? Is the theorem a sufficient justification?

• You can even change an infinite number of values. For example, $1_{[0,1]\setminus \mathbb{Q}}$ is the density of a uniform r.v. on $[0,1]$. – d.k.o. Jan 20 at 5:19
• Definition of density functions involves Lebesgue integration. If $f$ is a density function so is $g$ whenever $g$ is measurable and $f=g$ almost everywhere w.r.t. Lebesgue measure. Riemann integrability is not required for a function to be a density function. – Kabo Murphy Jan 20 at 5:23

Let's say we have some distribution over the real line in the sense that we have specified $$P(x\in A)$$ for all "sufficiently well behaved" $$A$$ and these probabilities satisfies a few axioms. We will say that $$P$$ is a continuous distribution if there is some $$f:\mathbb R \to \mathbb R^+$$ such that for any $$A$$ where the probability above is well defined, we have $$P(x \in A) = \int_A f(x)\,\mathrm dx$$ where the integral is the Lebesgue integral with respect to the Lebesgue measure on $$\mathbb R$$ (which you can think of as behaving exactly like the Reimann integral for most normal cases, but with some modifications to allow integration of a considerably larger class of functions). Any $$f$$ in the above definition will count as a "density". Now consider modifying $$f$$ at a finite set of points $$S$$. Formally, we can write this new density as $$g = f + h$$ where $$h = 0$$ everywhere except on $$S$$. Let's first integrate $$h$$. Even in the Reimann theory of integration, it is true that $$\int_A h(x)\,\mathrm dx = 0$$. This carries over to Lebesgue integration. Then for any $$A$$ that is "well behaved", we must have $$\int_A g(x)\,\mathrm dx = \int_A f(x) + g(x)\,\mathrm dx = \int_A f(x)\,\mathrm dx + \int g(x)\,\mathrm dx = \int_A f(x)\,\mathrm dx = P(x\in A)$$ But that means that $$g$$ also satisfies our definition of what it means to be a density. As mentioned by the comments before me, all of this can be generalized considerably. One way to put it is that modifying $$f$$ on any $$S$$ that satisfies $$\int 1\{S\}\,\mathrm dx = 0$$ will still give us a valid density (where $$1\{S\}$$ is the indicator function for the set $$S$$). Apologies if this explanation is not particularly satisfying. I am not sure how much measure theory knowledge to assume, so I am sweeping a ton under the rug. If you're interested in learning more about the formalization of probability, Billingsley's Probability and Measure is an excellent exposition.