A probability distribution and ordinary length obey the same structural rule: the size of a countable disjoint union is the sum of the sizes. Counting measure does too, assigning a finite set its number of points and an infinite set infinity. A measure is therefore not necessarily a probability and need not give a finite value to the whole space. The finiteness assumptions in later subtraction arguments must be stated explicitly.
Predict
Predict whether nested sets always have nested limiting sizes. For increasing sets the answer is yes, even when the limit is infinite. For decreasing sets a finite initial measure is needed. The tails of the positive integers decrease to the empty set, yet every tail has infinite counting measure. There is no valid cancellation of infinity minus infinity that would repair this example.
Experiment
Compare the two continuity directions.
Sets
Measure
Limit set
Limit of measures
[0,1−1/n]
Length
[0,1)
1
[0,1/n]
Length
Singleton 0
0
Integer tails from n onward
Counting
Empty
Infinity
The second row has finite starting measure. The third does not. Explain why the endpoint convention in the first row changes the union but not its length.
Understand
A measure μ on (X, F) is a map to[0,∞] with μ(empty)=0 and μ of a countable disjoint union equal to the sum. Monotonicity follows by writing B=A disjoint union (B minus A) when A⊆B. Finite additivity follows by appending empty sets. Countable subadditivity follows by disjointification: replace E_n by E_n minus all earlier E_j. These pieces are measurable, disjoint, and have the same union, while each has measure at most μ(E_n). Thus countable additivity supplies both exact and inequality-based tools.
Apply
If E_n increases, define D1=E1 and D_n=E_n minus E_(n−1). The D_n are disjoint, their first n union is E_n, and their full union is the increasing limit. Countable additivity equates the measure of that union with the limit of the partial sums μ(E_n). If E_n decreases and μ(E1)<∞, the complements inside E1 increase to E1 minus the intersection. Apply the increasing result and subtract from the finite μ(E1), obtaining continuity from above. The proof identifies precisely where finiteness enters. The same result holds if any later E_n has finite measure, after discarding the earlier finite prefix.
Continue
Finite measure means μ(X)<∞. Sigma-finite means X is a countable union of measurable sets of finite measure; ordinary length on the line is sigma-finite but not finite. A measure can be neither, as counting measure on an uncountable set shows. These distinctions will control uniqueness, product measures and density theorems. For each application, write the covering or finite set that supplies the hypothesis rather than assuming every familiar measure behaves like a probability. Continuity of sets is a consequence of countable additivity, not an extra topological continuity assumption.
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If μ(A)=4, μ(B)=5 and μ(A∩B)=2, find μ(A∪B).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For counting measure on positive integers, find μ({1,2,3,4}).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Prove continuity from below for increasing measurable E_n.
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Prove continuity from above assuming μ(E1)<∞.
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Give a decreasing sequence showing the finite hypothesis cannot be removed.
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Derive countable subadditivity from countable additivity.
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Reflect and continue
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