Measurable Set with Negative Measure has Negative Subset

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Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.

Let $A \in \Sigma$ have:

$-\infty < \map \mu A < 0$


Then:

there exists a $\mu$-negative set $B$ such that $B \subseteq A$ and $\map \mu B \le \map \mu A$.


Proof

Let:

$\delta_1 = \sup \set {\map \mu E : E \in \Sigma \text { and } E \subseteq A}$

possibly infinite.

Since $\O \in \Sigma$ and $\O \subseteq A$, we have:

$\map \mu \O = 0 \in \set {\map \mu E : E \in \Sigma \text { and } E \subseteq A}$

and so, from the definition of supremum, we have:

$\delta_1 \ge 0$.

Again applying the definition of supremum, we can pick $A_1 \in \Sigma$ so that:

$\ds \map \mu {A_1} \ge \min \set {\frac 1 2 \delta_1, 1} \ge 0$

We now define $\sequence {\delta_n}_{n \mathop \in \N}$ and $\sequence {A_n}_{n \mathop \in \N}$ recursively.

For $n > 1$, define:

$\ds \delta_n = \sup \set {\map \mu E : E \in \Sigma \text { and } E \subseteq A \setminus \bigcup_{i = 1}^{n - 1} A_i}$

Since:

$\ds \O \subseteq A \setminus \paren {\bigcup_{i = 1}^{n - 1} A_i}$

we have:

$\ds 0 \in \set {\map \mu E : E \in \Sigma \text { and } E \subseteq A \setminus \bigcup_{i = 1}^{n - 1} A_i}$

so, from the definition of supremum, we have:

$\delta_n \ge 0$

Now pick $A_n \in \Sigma$ so that:

$\ds A_n \subseteq A \setminus \paren {\bigcup_{i = 1}^{n - 1} A_i}$

and:

$\ds \map \mu {A_n} \ge \min \set {\frac 1 2 \delta_n, 1} \ge 0$

Since:

$\ds A_n \not \in \bigcup_{i = 1}^{n - 1} A_i$

we have that:

$A_n$ is disjoint to $A_1, A_2, \ldots, A_{n - 1}$


Now define:

$\ds A_\infty = \bigcup_{i = 1}^\infty A_i$

and $B = A \setminus A_\infty$.

As noted in the construction, we have that each $A_n$ is disjoint.

That is:

whenever $i \ne j$, we have $A_i \cap A_j = \O$.

So, since a signed measure is countably additive, we have:

$\ds \map \mu {A_\infty} = \sum_{i \mathop = 1}^\infty \map \mu {A_n}$

Since $\map \mu {A_n} \ge 0$ for each $n$, we therefore have:

$\map \mu {A_\infty} \ge 0$

We then have:

\(\ds \map \mu A\) \(=\) \(\ds \map \mu {A_\infty \cup \paren {A \setminus A_\infty} }\)
\(\ds \) \(=\) \(\ds \map \mu {A_\infty} + \map \mu {A \setminus A_\infty}\) since $\mu$ is countably additive
\(\ds \) \(=\) \(\ds \map \mu {A_\infty} + \map \mu B\) since $B = A \setminus A_\infty$
\(\ds \) \(\ge\) \(\ds \map \mu B\) since $\map \mu {A_\infty} \ge 0$


We now show that $B$ is $\mu$-negative.

Since $\map \mu A$ is finite, we have that:

$\map \mu {A_\infty}$ is finite.

So:

the series $\ds \sum_{i \mathop = 1}^\infty \map \mu {A_i}$ converges.

So, from Terms in Convergent Series Converge to Zero:

$\map \mu {A_n} \to 0$ as $n \to \infty$.

Since:

$\ds \map \mu {A_n} \ge \frac {\delta_n} 2$

we have:

$0 \le \delta_n \le 2 \map \mu {A_n}$

So:

$\delta_n \to 0$ as $n \to \infty$.

Let $E \in \Sigma$ have $E \subseteq B$.

We want to show that $\map \mu E \le 0$ so that $B$ is $\mu$-negative.

Recalling that $B = A \setminus A_\infty$, we have:

$\ds E \in \set {\map \mu E : E \in \Sigma \text { and } E \subseteq A \setminus \bigcup_{i = 1}^\infty A_i}$

Since we have:

$\ds A \setminus \bigcup_{i = 1}^\infty A_i \subseteq A \setminus \bigcup_{i = 1}^{n - 1} A_i$

we have the inclusion:

$\ds \set {\map \mu E : E \in \Sigma \text { and } E \subseteq A \setminus \bigcup_{i = 1}^\infty A_i} \subseteq \set {\map \mu E : E \in \Sigma \text { and } E \subseteq A \setminus \bigcup_{i = 1}^{n - 1} A_i}$

So, we obtain:

$\ds E \in \set {\map \mu E : E \in \Sigma \text { and } E \subseteq A \setminus \bigcup_{i = 1}^{n - 1} A_i}$

for each $n$.

From the definition of supremum, this gives:

$\map \mu E \le \delta_n$

for each $n$.

Taking $n \to \infty$, we obtain:

$\map \mu E \le 0$

as required.

$\blacksquare$


Sources

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