Monotone convergence theorem for non-negative numerical measurable functions

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Books disagree here, the version here may have a bias towards what I will be examined on


Let [ilmath](X,\mathcal{A},\mu)[/ilmath] be a measure space and let [ilmath](f_n)_{n\in\mathbb{N} }\subseteq\mathcal{M}\big(\mathcal{A},\mathcal{B}(\bar{\mathbb{R} }_{\ge 0})\big)[/ilmath][Note 1] be a sequence of measurable functions, [ilmath]f_n:X\rightarrow[0,+\infty]\eq\bar{\mathbb{R} }_{\ge 0} [/ilmath], then[1][2]:

  • if [ilmath]\forall n\in\mathbb{N}[f_n\le f_{n+1}][/ilmath][Note 2] - i.e. [ilmath](f_n)_{n\in\mathbb{N} } [/ilmath] is a non-decreasing sequence - then:
    • [math]\int \mathop{\text{lim} }_{n\rightarrow\infty} \Big(f_n\Big)\ \mathrm{d}\mu\eq\mathop{\text{lim} }_{n\rightarrow\infty}\left(\int f_n\ \mathrm{d}\mu\right) [/math][Note 3]

This could be phrased differently; as an alternative statement:

  • Define [ilmath]f:X\rightarrow[0,+\infty][/ilmath] by [math]f:x\mapsto\mathop{\text{lim} }_{n\rightarrow\infty}\Big(f_n(x)\Big)[/math], this limit exists forall [ilmath]x\in X[/ilmath] as we allow the value [ilmath]+\infty[/ilmath].
    • Then we have:
      1. [ilmath]f\in\mathcal{M}\big(\mathcal{A},\mathcal{B}(\bar{\mathbb{R} }_{\ge 0})\big)[/ilmath][Todo 1] - [ilmath]f[/ilmath] is a measurable function itself - and
      2. [math]\int f\ \mathrm{d}\mu\eq\mathop{\text{lim} }_{n\rightarrow\infty}\left(\int f_n\ \mathrm{d}\mu\right) [/math]


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Done on paper? Find it and post here? Do again? Can be found in[1] before claim 9.11 which is (a form of) Fatou's lemma


  1. Suppose we had:
    1. [ilmath]f\in\mathcal{M}\big(\mathcal{A},\mathcal{B}(\bar{\mathbb{R} }_{\ge 0})\big)[/ilmath] and
    2. [ilmath]f\in\mathcal{L}^+:\eq\left\{f:X\rightarrow\bar{\mathbb{R} }\ \Big\vert\ \forall x\in X[0\le f(x)]\wedge f\in\mathcal{M}\big(\mathcal{A},\mathcal{B}(\bar{\mathbb{R} })\big)\right\} [/ilmath]
    Then these are the same requirements for [ilmath]f[/ilmath], that is [ilmath]\mathcal{M}\big(\mathcal{A},\mathcal{B}(\bar{\mathbb{R} }_{\ge 0})\big)\eq\mathcal{L}^+ [/ilmath]
    • I've seen (but not read) a proof and trust the source - Alec -17/April/2017 - 0908
  2. [ilmath]f\le g[/ilmath] is short for [ilmath]\forall x\in X[f(x)\le g(x)][/ilmath]
  3. Note that for the integral of a non-negative numerical measurable function to be even defined that (as the name suggests) the function must be a measurable function. This is covered in the "alternative statement".


  1. 1.0 1.1 Measures, Integrals and Martingales - René L. Schilling
  2. A Guide To Advanced Real Analysis - Gerald B. Folland


  1. TODO: Link to specific proof!