Split density mass from singular mass

Try

A distribution combines a continuous background with a concentrated point charge. Relative to length on [0,1], let ν have density three plus a point mass of size two at zero. Its total is five, but only three units can be described by an ordinary length density. The remaining two units sit on a length-null set.

Predict

Predict whether every singular measure must consist of point masses. It need not. The Cantor distribution is concentrated on a null set yet gives every individual point mass zero. Thus continuous distribution functions and length densities are different concepts: the absence of jumps does not guarantee an absolutely continuous law.

Experiment

Compare three components relative to length.

Measure	Has length density	Has atoms	Concentrated on a null set
Uniform law on [0,1]	Yes	No	No
Point mass at zero	No	Yes	Yes
Cantor distribution	No	No	Yes

For the mixture in the opening example, calculate its mass on the singleton zero and on the interval (0,1/2]. Explain why integrating the background density alone misses the singleton mass.

Understand

Two positive measures are mutually singular when they are concentrated on disjoint measurable sets. The Lebesgue decomposition theorem for sigma-finite positive measures writes ν uniquely as ν_ac+ν_s, where ν_ac≪μ and ν_s is singular to μ. The density of ν_ac is its Radon–Nikodym derivative; the singular part has no such density unless it is zero.

For finite measures set λ=μ+ν. Radon–Nikodym gives h=dμ/dλ and g=dν/dλ. Let A be the measurable set where h>0. On A, define f=g/h; put it zero elsewhere. The identity for integrating with a density shows integral_E f dμ=integral_(E intersection A) g dλ=ν(E intersection A). This is the absolutely continuous component. Define ν_s(E)=ν(E intersection A complement). Since μ(A complement)=integral_(h=0)h dλ=0, this second component is singular. Their sum is ν. For sigma-finite measures, perform the construction on disjoint pieces where both are finite and combine them.

For uniqueness, suppose two decompositions have singular supports N_1 and N_2, each μ-null. On the complement of their union, both absolutely continuous parts equal ν. On the union, both vanish by absolute continuity. Thus the absolutely continuous measures agree on every set, and subtracting them from ν gives equality of the singular measures.

To construct a singular nonatomic example, take binary digits b_k of a uniform point in [0,1], choosing a fixed convention at the countably many ambiguous endpoints. Map it to Σ2b_k/3^k. The map is measurable as a limit of simple functions. Its image lies in the ternary Cantor set, whose nth-stage length is (2/3)^n and whose length is zero by continuity above. A specified output fixes the digit choices, up to endpoint conventions, and therefore has an input preimage of length zero. The pushforward is a probability measure with no atoms but singular to length.

Apply

Change the reference measure to length plus a point mass at zero. The earlier point charge now has a density relative to this new reference. Identify a density that equals two at zero and three away from zero for the opening mixture. This exercise shows why every density statement must name its reference measure.

Continue

The decomposition separates two ways mass can be distributed, rather than ranking measures as regular or defective. Next, differentiation of averages connects an absolutely continuous density to local observations. Singular mass requires separate treatment and cannot be recovered by blindly differentiating an ordinary density formula.

Continue to Recover an integrable function from local averages.