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The interval data on $(a, b]$ determines the Borel measure uniquely

Statement

Let $μ$ and $ν$ be Borel measures on $R$ finite on compact sets in the sense of A Borel measure on $R$ that is finite on compact sets. If

$μ ((a, b]) = ν ((a, b]) for every a < b,$

then $μ (E) = ν (E)$ for every Borel set $E \subseteq R$ .

Facts & Assumptions

Given: Two Borel measures $μ, ν$ on $R$ , each finite on compact sets, and agreement of $μ$ and $ν$ on every half-open interval $(a, b]$ .

[L1]

The family of half-open intervals $(a, b]$ with $a < b$ generates the Borel sigma-algebra on $R$ . (Seven generating families for the Borel sigma-algebra on the real line)

[L2]

Measures that agree on a generating pi-system and on an increasing finite-measure exhaustion from that pi-system agree on the whole sigma-algebra. (Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system)

Proof

technique · direct

1.1

Let

L1algebra

$P : = {(a, b] : a < b} .$

If $I_{1} = (a, b]$ and $I_{2} = (c, d]$ lie in $P$ , then $I_{1} \cap I_{2}$ is either empty or another half-open interval $(\max {a, c}, \min {b, d}]$ , so $P$ is a pi-system. By [L1], $σ (P) = B (R)$ . [L1, algebra]

1.2

For each $n \in N$ , put $P_{n} : = (- n - 1, n] \in P$ . The [given, algebra] sequence $(P_{n})$ is increasing and $⋃_{n} P_{n} = R$ . Because each $P_{n}$ is contained in the compact interval $[- n - 1, n]$ , both $μ (P_{n})$ and $ν (P_{n})$ are finite; and by the hypothesis they are equal.

givenalgebra

2.1

Step 1.1 provides the generating pi-system and step 1.2 provides the [step 1.1, step 1.2, L2] increasing finite-measure exhaustion. Therefore [L2] applies and yields $μ = ν$ on $B (R)$ .

step 1.1step 1.2L2∎

Depends on

Used by

Dependency tree · two levels

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., Theorem 1.16 (standard reference, not scraped)

The interval data on (a,b](a,b] determines the Borel measure uniquely