关于Lebesgue Decomposition Theorem中discrete compound存在性的证明求助

Hey there! No worries about your English at all—totally get where you're coming from with this tricky detail. The three-part Lebesgue decomposition (discrete + absolutely continuous + singular continuous) is often presented without explicitly walking through the discrete component, so let's break that down clearly:

First, let's formalize what a discrete measure is #

A discrete measure (sometimes called an atomic measure) is one that can be written as a countable sum of point masses: there exists a countable set of points ${x_n}{n=1}^\infty$ and positive real numbers ${a_n}{n=1}^\infty$ such that for every measurable set $E$,
$$\mu_d(E) = \sum_{x_n \in E} a_n$$

Step 1: Identify all "atomic" points for your measure $\mu$ #

Let $S = { x \in X \mid \mu({x}) > 0 }$, where $X$ is your underlying measure space. We first need to show $S$ is countable:

• For each positive integer $k$, define $S_k = { x \mid \mu({x}) \geq 1/k }$.
• Each $S_k$ must be finite: if $S_k$ were infinite, we could pick an infinite sequence of distinct points from it, and the measure of that sequence would be at least $\sum_{n=1}^\infty 1/k = \infty$, which contradicts $\mu$ being $\sigma$-finite (a standard assumption for Lebesgue decomposition; if $\mu$ is finite, this is even more straightforward).
• Since $S = \bigcup_{k=1}^\infty S_k$, it's a countable union of finite sets—hence countable.

Step 2: Define the discrete component directly #

Set $\mu_d(E) = \mu(E \cap S)$ for all measurable $E$. Since $S$ is countable, this is exactly a sum of point masses: each $x \in S$ contributes $\mu({x})$ to the measure of any set containing $x$, so $\mu_d$ fits the definition of a discrete measure.

Step 3: Confirm the remaining measure has no atoms #

Let $\mu' = \mu - \mu_d$. For any single point $x$:

• If $x \in S$, then $\mu'({x}) = \mu({x}) - \mu_d({x}) = \mu({x}) - \mu({x}) = 0$.
• If $x \notin S$, then $\mu({x}) = 0$, so $\mu'({x}) = 0 - 0 = 0$.

This means $\mu'$ has no atomic points, so you can then apply the standard two-part Lebesgue decomposition to $\mu'$ to split it into the absolutely continuous part $\mu_{ac}$ (with respect to your reference measure, usually Lebesgue measure) and the singular continuous part $\mu_{sc}$ (which is singular with respect to the reference measure and has no atoms).

Putting it all together, you get the three-part decomposition:
$$\mu = \mu_d + \mu_{ac} + \mu_{sc}$$

A quick side note: Sometimes textbooks lump the discrete measure into the "singular" category (since its support is a countable set, which has Lebesgue measure zero), but the three-part version separates it because its structure is fundamentally different from singular continuous measures (which live on uncountable null sets).

备注：内容来源于stack exchange，提问作者Kiko