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Signed Measures, Absolute Continuity, and Radon–Nikodym

The capstone. Allow measures to take negative values, decompose them, and ask when one measure is a 'weighted version' of another. The Radon–Nikodym theorem answers it: absolute continuity is exactly the existence of a density. We finish with convergence in measure versus in norm.

Signed measures

A signed measure ν on (X, M) is countably additive but allowed to take values in (−∞, +∞] (or [−∞, +∞)), with ν(∅) = 0. The model example is ν(E) = ∫_E g dμ for an integrable g that changes sign. The Hahn decomposition splits X into a positive set P and a negative set N, and the Jordan decomposition writes ν = ν⁺ − ν⁻ as a difference of two genuine (non-negative) measures living on P and N.

Absolute continuity and the density

A measure ν is absolutely continuous with respect to μ, written ν ≪ μ, if every μ-null set is also ν-null: μ(E) = 0 ⇒ ν(E) = 0. Intuitively ν cannot 'see' anything μ ignores. The Radon–Nikodym theorem says that for σ-finite measures this is equivalent to ν being given by a density: there is a non-negative measurable g with ν(E) = ∫_E g dμ for all E. That g is the Radon–Nikodym derivative dν/dμ, unique up to almost-everywhere μ-equality.

Easy direction: density ⇒ absolute continuity.

Suppose ν(E) = ∫_E g dμ  with g ≥ 0 measurable.
Let E satisfy μ(E) = 0.
Then the integrand g·1_E is 0 μ-almost everywhere,
since it can be nonzero only on E, a μ-null set. Hence
        ν(E) = ∫_E g dμ = ∫ g·1_E dμ = 0.
So μ(E) = 0 ⇒ ν(E) = 0, i.e. ν ≪ μ. ∎

Uniqueness of the density.
If ∫_E g₁ dμ = ∫_E g₂ dμ for all E, take
        E = { g₁ > g₂ }.  Then ∫_E (g₁ − g₂) dμ = 0
with integrand ≥ 0, forcing μ(E) = 0; symmetrically μ{g₂ > g₁}=0.
Thus g₁ = g₂ μ-a.e. ∎
Density ⇒ absolute continuity is one line; the density is a.e.-unique.

Convergence in measure versus in norm

A sequence fₙ → f in measure if for every ε > 0, μ{ |fₙ − f| > ε } → 0. This is genuinely weaker than Lᵖ-norm convergence: norm convergence implies convergence in measure (by Chebyshev), but not the reverse. The traveling-bump examples below show how each notion can hold while the others fail.

Chebyshev: norm convergence ⇒ convergence in measure.
For ε > 0, on the set A = { |fₙ − f| > ε } we have |fₙ−f|ᵖ > εᵖ, so
    εᵖ · μ(A) ≤ ∫_A |fₙ−f|ᵖ dμ ≤ ‖fₙ − f‖ₚᵖ.
Thus  μ{ |fₙ−f| > ε } ≤ ‖fₙ − f‖ₚᵖ / εᵖ → 0.

Converse fails — the 'sliding bump' on [0,1] (Lebesgue):
  enumerate dyadic intervals I = [j/2ᵏ, (j+1)/2ᵏ] in order of length,
  and let fₙ = 1_{Iₙ}.  The bump width → 0, so for 0 < ε < 1:
      μ{ fₙ > ε } = length(Iₙ) → 0   (converges in measure to 0),
  yet ‖fₙ‖ₚ = length(Iₙ)^(1/p) → 0 here too — refine instead with
      gₙ = 2^{k/p} · 1_{Iₙ}.  Then ‖gₙ‖ₚ = 1 for all n (no norm limit),
  while still gₙ → 0 in measure. So in measure ⇏ in norm. ∎
Chebyshev gives one implication; a height-boosted sliding bump kills the converse.