Editing Radon–Nikodym theorem (section)

===For finite measures===

'''Constructing an extended-valued candidate'''  First, suppose {{mvar|μ}} and {{mvar|ν}} are both finite-valued nonnegative measures. Let {{mvar|F}} be the set of those extended-value measurable functions {{math|''f''  : ''X'' → [0, ∞]}} such that:
:<math>\forall A \in \Sigma:\qquad \int_A f\,d\mu \leq \nu(A)</math>

{{math|''F'' ≠ ∅}}, since it contains at least the zero function. Now let {{math|''f''<sub>1</sub>,  ''f''<sub>2</sub> ∈ ''F''}}, and suppose {{mvar|A}} is an arbitrary measurable set, and define:
:<math>\begin{align}
  A_1 &= \left\{ x \in A : f_1(x) > f_2(x) \right\}, \\
  A_2 &= \left\{ x \in A : f_2(x) \geq f_1(x) \right\}.
\end{align}</math>

Then one has
:<math>\int_A\max\left\{f_1, f_2\right\}\,d\mu = \int_{A_1} f_1\,d\mu + \int_{A_2} f_2\,d\mu \leq \nu\left(A_1\right) + \nu\left(A_2\right) = \nu(A),</math>

and therefore, {{math|max{ ''f'' <sub>1</sub>,  ''f'' <sub>2</sub>} ∈ ''F''}}.

Now, let {{math|{{mset| ''f<sub>n</sub>'' }}}} be a sequence of functions in {{mvar|F}} such that
:<math>\lim_{n\to\infty}\int_X f_n\,d\mu = \sup_{f\in F} \int_X f\,d\mu.</math>

By replacing {{math| ''f<sub>n</sub>'' }} with the maximum of the first {{mvar|n}} functions, one can assume that the sequence {{math|{{mset| ''f<sub>n</sub>'' }}}} is increasing. Let {{mvar|g}} be an extended-valued function defined as
:<math>g(x) := \lim_{n\to\infty}f_n(x).</math>

By Lebesgue's [[monotone convergence theorem]], one has
:<math>\lim_{n\to\infty} \int_A f_n\,d\mu = \int_A \lim_{n\to\infty} f_n(x)\,d\mu(x) = \int_A g\,d\mu \leq \nu(A)</math>

for each {{math|''A'' ∈ Σ}}, and hence, {{math|''g'' ∈ ''F''}}. Also, by the construction of {{mvar|g}},
:<math>\int_X g\,d\mu = \sup_{f\in F}\int_X f\,d\mu.</math>

'''Proving equality'''  Now, since {{math|''g'' ∈ ''F''}},
:<math>\nu_0(A) := \nu(A) - \int_A g\,d\mu</math>

defines a nonnegative measure on {{math|Σ}}. To prove equality, we show that {{math|1=''ν''<sub>0</sub> = 0}}.

Suppose {{math|ν<sub>0</sub> ≠ 0}}; then, since {{mvar|μ}} is finite, there is an {{math|''ε'' > 0}} such that {{math|''ν''<sub>0</sub>(''X'') > ''ε μ''(''X'')}}. To derive a contradiction from {{math|ν<sub>0</sub> ≠ 0}}, we look for a [[Positive and negative sets|positive set]] {{math|''P'' ∈ Σ}} for the signed measure {{math|''ν''<sub>0</sub> − ''ε μ''}} (i.e. a measurable set {{mvar|P}}, all of whose measurable subsets have non-negative {{math|''ν''<sub>0</sub>&nbsp;−&nbsp;''εμ''}} measure), where also {{mvar|P}} has positive {{mvar|μ}}-measure. Conceptually, we're looking for a set {{mvar|P}}, where {{math|''ν''<sub>0</sub> ≥ ''ε μ''}} in every part of {{mvar|P}}. A convenient approach is to use the [[Hahn decomposition theorem|Hahn decomposition]] {{math|(''P'', ''N'')}} for the signed measure {{math|''ν''<sub>0</sub> − ''ε μ''}}.

Note then that for every {{math|''A'' ∈ Σ}} one has {{math|''ν''<sub>0</sub>(''A'' ∩ ''P'') ≥ ''ε μ''(''A'' ∩ ''P'')}}, and hence,

:<math>\begin{align}
  \nu(A) &=    \int_A g\,d\mu + \nu_0(A) \\
         &\geq \int_A g\,d\mu + \nu_0(A\cap P)\\
         &\geq \int_A g\,d\mu + \varepsilon\mu(A\cap P)
          =    \int_A\left(g + \varepsilon 1_P\right)\,d\mu,
\end{align}</math>

where {{math|1<sub>''P''</sub>}} is the [[indicator function]] of {{mvar|P}}. Also, note that {{math|''μ''(''P'') > 0}} as desired; for if {{math|1=''μ''(''P'') = 0}}, then (since {{mvar|ν}} is absolutely continuous in relation to {{mvar|μ}}) {{math|1=''ν''<sub>0</sub>(''P'') ≤ ''ν''(''P'') = 0}}, so {{math|1=''ν''<sub>0</sub>(''P'') = 0}} and
:<math>\nu_0(X) - \varepsilon\mu(X) = \left(\nu_0 - \varepsilon\mu\right)(N) \leq 0,</math>
contradicting the fact that {{math|''ν''<sub>0</sub>(''X'') > ''εμ''(''X'')}}.

Then, since also
:<math>\int_X\left(g + \varepsilon1_P\right)\,d\mu \leq \nu(X) < +\infty,</math>

{{math|''g'' + ''ε'' 1<sub>''P''</sub> ∈ ''F''}} and satisfies
:<math>\int_X\left(g + \varepsilon 1_P\right)\,d\mu > \int_X g\,d\mu = \sup_{f\in F}\int_X f\,d\mu.</math>

This is [[reductio ad absurdum|impossible]] because it violates the definition of a [[supremum]]; therefore, the initial assumption that {{math|''ν''<sub>0</sub> ≠ 0}} must be false. Hence, {{math|1=''ν''<sub>0</sub> = 0}}, as desired.

'''Restricting to finite values'''  Now, since {{mvar|g}} is {{mvar|μ}}-integrable, the set {{math|{{mset|1=''x'' ∈ ''X'' : ''g''(''x'') = ∞}}}} is {{mvar|μ}}-[[null set|null]]. Therefore, if a {{math| ''f'' }} is defined as
:<math>f(x) = \begin{cases}
  g(x) & \text{if }g(x) < \infty \\
  0    & \text{otherwise,}
\end{cases}</math>

then {{math|''f''}} has the desired properties.

'''Uniqueness'''  As for the uniqueness, let {{math| ''f'', ''g'' : ''X'' → [0, ∞)}} be measurable functions satisfying
:<math>\nu(A) = \int_A f\,d\mu = \int_A g\,d\mu</math>

for every measurable set {{mvar|A}}. Then, {{math|''g'' − ''f'' }} is {{mvar|μ}}-integrable, and
:<math>\int_A(g - f)\,d\mu = 0.</math> (Recall that we can split the integral into two as long as they are measurable and non-negative)

In particular, for {{math|1=''A'' = {''x'' ∈ ''X'' : ''f''(''x'') > ''g''(''x'')},}} or {{math|{{mset|''x'' ∈ ''X'' : ''f''(''x'') < ''g''(''x'')}}}}. It follows that
:<math>\int_X(g - f)^+\,d\mu = 0 = \int_X(g - f)^-\,d\mu,</math>

and so, that {{math|1=(''g'' − ''f'' )<sup>+</sup> = 0}} {{mvar|μ}}-almost everywhere; the same is true for {{math|(''g'' − ''f'' )<sup>−</sup>}}, and thus, {{math|1=''f'' = ''g''}} {{mvar|μ}}-almost everywhere, as desired.