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A step function generates a finite atomic measure

Example

Fix numbers $x_{1} < \dots < x_{m}$ and positive masses $α_{1}, \dots, α_{m}$ , and define

$F (x) : = \sum_{x_{j} \leq x} α_{j} .$

Then the Lebesgue-Stieltjes measure of $F$ is the finite atomic measure

$μ_{F} = \sum_{j = 1}^{m} α_{j} δ_{x_{j}} .$

Facts & Assumptions

Given: Points $x_{1} < \dots < x_{m}$ , positive numbers $α_{1}, \dots, α_{m}$ , the step function $F (x) = \sum_{x_{j} \leq x} α_{j}$ , and its Lebesgue-Stieltjes measure $μ_{F}$ .

[L1]

Finite weighted sums of Dirac measures are measures. (A Dirac set function is a probability measure, Nonnegative scalar multiples and countable weighted sums of measures are measures)

[L2]

A Borel measure on $R$ finite on compact sets is uniquely determined by its values on half-open intervals. (The interval data on $(a, b]$ determines the Borel measure uniquely)

Verification

technique · direct

1.1L1

Let $ν : = \sum_{j = 1}^{m} α_{j} δ_{x_{j}}$ . By [L1], this is a Borel measure on $R$ .

2.1givenstep 1.1algebra

For every $a < b$ ,

$ν ((a, b]) = \sum_{a < x_{j} \leq b} α_{j} .$

On the other hand, because $F (x) = \sum_{x_{j} \leq x} α_{j}$ , one has

$F (b) - F (a) = \sum_{x_{j} \leq b} α_{j} - \sum_{x_{j} \leq a} α_{j} = \sum_{a < x_{j} \leq b} α_{j} = ν ((a, b]) .$

So $μ_{F}$ and $ν$ agree on every half-open interval $(a, b]$ . [given, step 1.1, algebra]

3.1step 2.1L2∎

Both $μ_{F}$ and $ν$ are Borel measures on $R$ finite on [step 2.1, L2] compact sets. By step 2.1 and [L2], they are equal on every Borel set. Thus $μ_{F} = \sum_{j = 1}^{m} α_{j} δ_{x_{j}}$ , which is the claimed formula.

Depends on

Used by

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Dependency tree · two levels

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., Section 1.5 (standard reference, not scraped)