Lebesgue decomposition and Radon-Nikodym derivative given a function.

Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by $$f(x)=\begin{cases}0 & \text{ if }-\infty<x<0\\ 1 &\text{ if }0\leq x <1\\ x^{2}+x^{3} &\text{ if }1\leq x <2\\ 17 &\text{ if }2\leq x<\infty.\end{cases}$$ Since $f$ is bounded, monotonically increasing and right-continuous, we know that it gives rise to a finite, positive Borel measure $\mu$ whose cumulative distribution function is $f$. Determine the Lebesgue decomposition of $\mu$ relative to Lebesgue measure, $m$, and determine the Radon-Nikodym derivative of $\mu_{a}$ with respect to $m$, where $\mu_{a}$ denotes the absolutely continuous part of the Lebesgue decomposition.

It has been a while since I worked with these ideas, and I didn't entirely understand them when we were covering them, so any help is appreciated.

I think we define the measure by $$\mu(E)=\int_{E}f'dm.$$ Is this correct? If so, I'm not sure how to proceed from here.

We have $$f(x) = \chi_{[0,1)}(x) + x^2(1+x)\chi_{[1,2)}(x) + 17\chi_{[2,\infty)}(x).$$ The Lebesgue-Stieltjes measure associated with $f$ is defined by $$\mu(E) = \int_E \mathsf df, \ E\in\mathcal B(\mathbb R)$$ where the integral is interpreted as a Stieltjes integral.

Note that $f$ has three jump discontinuities: \begin{align} \Delta_0 := f(0)-f(0-) &= 1-0 = 1\\ \Delta_1 := f(1)-f(1-) &= 2-1 = 1\\ \Delta_2 := f(2)-f(2-) &= 17-12 = 5. \end{align} It follows that the Lebesgue decomposition of $\mu$ is $\mu = \mu_a + \mu_d$ where $$\mu_a(E) = \int_{E\ \cap\ (1,2)} f'(x) \mathsf dx,$$ and $$\mu_d(E) = \Delta_0\chi_E(0) + \Delta_1\chi_E(1) + \Delta_2\chi_E(2)$$ are the absolutely continuous and discrete parts of $\mu$, respectively. It is clear that $\mu_a\perp\mu_d$ and $\mu_a\ll m$.

The distribution function of $\mu$ is $F(x) = \mu((-\infty,x])$. $F$ is a càdlàg function, and is similar in nature to a cumulative distribution function of a random variable. In fact, since $F\geqslant 0$ and

\begin{align} \lim_{x\to\infty}F(x) &= (F(0) - F(0-)) + (F(1) - F(1-))\\ &\quad+ (F(2-) - F(1)) + (F(2) - F(2-))\\ &= \Delta_0 +\Delta_1 + \int_1^2 (2x+3x^2)\,\mathsf dx +\Delta_2\\ &= 1 + 1 + 10 + 5\\ &= 17, \end{align} we see that $\nu:=\frac1{17}\mu$ is a probability measure on $\mathbb R$.

Edit: A technical detail. The condition $\mu=\mu_a+\nu$ where $\mu_a\ll m$ and $\mu_a\perp\nu$ does not imply that $\nu$ is a discrete measure, i.e. $\nu$ need not be concentraed on a countable set. In full generality, $\nu = \nu_d + \nu_{sc}$ where $\nu_d$ is a discrete measure and $\nu_{sc}$ is a singular measure. That is, $\nu_{sc}$ is concentrated on a set of Lebesgue measure zero, but $\nu_{sc}(\{x\})=0$ for all $x\in\mathbb R$. In this example, $\mu$ has no singular part, since the set on which $F$ is not absolutely continuous is precisely the set of point masses of $\mu$.

• Every measure on B(R) has three parts: one part with density, one part made of Dirac masses and a third part, which is singular with no atoms. Your answer could make believe that this third part does not exist in general. – Did Jul 23 '16 at 16:29
• @Math1000 This makes sense! Thank you – Michael Shumate Jul 23 '16 at 16:41
• @Did I am well aware, and if this measure had a singular part I would have been sure to include that in my answer. – Math1000 Jul 23 '16 at 17:05
• Then why is there no argument in your post to show the singular part does not exist in the present case? – Did Jul 23 '16 at 17:11

The absolute continuous part is easy to find. By graphing the distribution function you can easily find that there are three points that "jump",x=0 with jump 1, x=1 with jump 1, and x=2 with jump 5. So the singular part is 1*delta0 + 1*delta1 + 5*delta2 delta0 means Dirac measure with unit mass on point x=0, delta1 and delta2 are similar.