Integration 


    
    
      
        
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          Show something holds for indicator functions.
        
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          Show it holds for simple functions by linearity.
        
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          Use \(s_k \nearrow f\) and apply MCT to show it holds for \(f\).
        
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          \(L^+\): nonnegative measurable functions
        
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          \(L^1\): Lebesgue integrable functions, so \({\left\lVert {f} \right\rVert}_{L^1} \coloneqq\int {\left\lvert {f} \right\rvert} < \infty\).
        
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      A function \(f: (X, {\mathcal{M}}_X) \to (Y, {\mathcal{M}}_Y)\) is \(({\mathcal{M}}_X, {\mathcal{M}}_Y){\hbox{-}}\)measurable iff \(f^{-1}({\mathcal{M}}_Y) \subseteq {\mathcal{M}}_X\). Equivalently, if \({\mathcal{E}}_Y\) is a generating set for \({\mathcal{B}}_Y\), \(f^{-1}({\mathcal{E}}_Y) \subseteq {\mathcal{B}}_X\).
    

  
    
    
      
        
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          An functional on a general measurable space \(f: f:(X, {\mathcal{M}}_X) \to ({\mathbf{R}}, {\mathcal{B}}_{\mathbf{R}})\) is measurable \(\iff f\) is \(({\mathcal{M}}_X, {\mathcal{B}}_{\mathbf{R}}){\hbox{-}}\)measurable.
        
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          A functional \(f: {\mathbf{R}}^d\to {\mathbf{R}}\) is Borel measurable iff \(f\) is \(({\mathcal{B}}_{{\mathbf{R}}^d}, {\mathcal{B}}_{\mathbf{R}}){\hbox{-}}\)measurable.
        
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          A functional \(f: {\mathbf{R}}^d\to {\mathbf{R}}\) is Lebesgue measurable iff \(f\) is \(({\mathcal{L}}_{{\mathbf{R}}^d}, {\mathcal{B}}_{\mathbf{R}}){\hbox{-}}\)measurable.
        
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      Using that \({\mathcal{B}}_{{\mathbf{R}}}\) is generated by open/closed rays, it suffices to check any of the following (for all \(\alpha \in {\mathbf{R}}\)):
    

  
    
    
      
        
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          \(f^{-1}(\alpha, \infty)\in {\mathcal{M}}\)
        
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          \(f^{-1}[\alpha, \infty)\in {\mathcal{M}}\)
        
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          \(f^{-1}(-\infty, \alpha)\in {\mathcal{M}}\)
        
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          \(f^{-1}(-\infty, \alpha]\in {\mathcal{M}}\)
        
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      Note that we still require Borel sets in the target for Lebesgue measurability! Taking \(({\mathcal{L}}_{{\mathbf{R}}^d}, {\mathcal{L}}_{\mathbf{R}})\) functions is too stringent, e.g. this class does not contain continuous functionals.
    

  

    

      If \(f\) is \({\mathcal{L}}{\hbox{-}}\)measurable and \(h\) is continuous, it’s not necessarily true that \(k\coloneqq f\circ h\) is \({\mathcal{L}}{\hbox{-}}\)measurable. Standard counterexample: set \(g(x) \coloneqq C(x) + x\) for \(C\) the Cantor-Lebesgue function, then \(g:[0, 1]\to [0, 2]\) is a homeomorphism. Then \(m(g(C)) = 1\) since \(f\) is constant on intervals in \(C^c\), so use Vitali’s theorem: a set is null iff every subset is measurable. So \(g(C)\) contains a non-measurable set \(A\). Define \(B\coloneqq g^{-1}(A)\), then \(B \subset C\) and \(m(C) = 0\) implies \(B\) is measurable and \(\chi_B\) is a measurable function. But then \(k\coloneqq\chi_B \circ g^{-1}\) is not \({\mathcal{L}}{\hbox{-}}\)measurable, since \(k^{-1}(1) = A\) is a non-measurable set, but \(\chi_B\) is \({\mathcal{L}}{\hbox{-}}\)measurable and \(g^{-1}\) is continuous.
    

  

    

      \({\mathcal{M}}{\hbox{-}}\)measurable functionals are closed under
    

  
    
    
      
        
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          Sums
        
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          Products
        
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          Sups/infs
        
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          Limsups/Liminfs
        
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          Limits when they exist, and the limiting function is measurable.
        
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          \(\max(f, g)\) and \(\min(f, g)\).
        
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      Characteristic functions on measurable sets are automatically measurable, since \(E\in {\mathcal{M}}\implies E = \chi_E^{-1}(\left\{{1}\right\})\).
    

  

    

      A simple function \(s: {\mathbf{C}}\to X\) is a finite linear combination of indicator functions of measurable sets, i.e.  

\begin{align*}
s(x) = \sum_{j=1}^n c_j \chi_{E_j}(x)
.\end{align*}

    

  

    

      If \(f:X\to {\mathbf{C}}\) is measurable, there is a sequence of simple functions \(\phi_n\nearrow f\) which always converges pointwise, and converges uniformly on any bounded set.
    

  

    

      

\begin{align*}
\int_X f \coloneqq\sup \left\{{ \int s(x) \,d\mu{~\mathrel{\Big\vert}~}0\leq s \leq f, s\text{ simple } }\right\} 
.\end{align*}

    

  
    

      Note that if \(s = \sum c_j \chi_{E_j}\) is simple, then 

\begin{align*}
\int_X s(x) \,d\mu\coloneqq\sum_{j=1}^n c_j \mu(E_j)
.\end{align*}

    

  

    

      A useful fact: for \((X, \mathcal{M})\) a measure space, integrals split across disjoint sets: 

\begin{align*}
\int_X f = \int_{X\setminus A} f + \int_A f && \forall\, A \in \mathcal{M} 
.\end{align*}

    

  

    

      An essential lower bound \(b\) on a function \(f\) is any real number such that \(S_{b} \coloneqq\left\{{x{~\mathrel{\Big\vert}~}f(x) < b }\right\} = f^{-1}(-\infty, b)\) has measure zero. The essential infimum is the supremum of all essential lower bounds, i.e. \({\mathrm{ess}}\inf f \coloneqq\sup_{b} \left\{{b{~\mathrel{\Big\vert}~}\mu S_b = 0}\right\}\). This is the greatest lower bound almost everywhere.
    

  
    

      Similarly an essential upper bound \(c\) is any number such that \(S^c \coloneqq f^{-1}(c, \infty)\) has measure zero, and the essential supremum is \({\mathrm{ess}}\sup f \coloneqq\inf_{c} \left\{{c{~\mathrel{\Big\vert}~}\mu S^c = 0}\right\}\), which is the least upper bound almost everywhere.
    

  
    

      A function is essentially bounded if \({\left\lVert {f} \right\rVert}_\infty \coloneqq{\mathrm{ess}}\sup f < \infty\). These are functions which are bounded almost everywhere.
    

  

    

      \(f(x) = x\chi_{\mathbf{Q}}(x)\) is essentially bounded but not bounded.
    

  

    

      If \(f\in L^\infty(X)\), then \(f\) is equal to some bounded function \(g\) almost everywhere.
    

  

    

      

\begin{align*}
\int_0^1 {1\over x^p} < \infty \iff  p < 1 \\
\int_1^\infty {1\over x^p} < \infty \iff  p > 1 
.\end{align*}

    

  

    

      Large powers of \(x\) help us in neighborhoods of infinity and hurt around zero.
    

  
The Convergence Theorems

    

      If \(f_n: X\to [0, \infty) \in L^+\) and \(f_n \nearrow f\) almost everywhere, then 

\begin{align*}
\lim \int f_n
= \int \lim f_n = \int f
\quad \text{i.e.}~~ \int f_n \to \int f
.\end{align*}

    

  

    

      Measurable, non-negative, increasing pointwise a.e. allows commuting limits and integrals.
    

  

    

      #todo Proof
    

  

    

      If \(f_n \in L^1\) and \(f_n \to f\) almost everywhere with \({\left\lvert {f_n} \right\rvert} \leq g\) for some \(g\in L^1\), then \(f\in L^1\) and 

\begin{align*}
\int {\left\lvert {f_n - f} \right\rvert} \to 0
.\end{align*}

    

  
    

      As a consequence, 

\begin{align*}
\lim \int f_n = \int \lim f_n = \int f \quad \text{i.e.}~~ \int f_n \to \int f < \infty
\end{align*}

    

  
    

      
    

      Positivity not needed.
    

  
    

  

    

      #todo Proof
    

  

    

      If
    

  
    
    
      
        
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          \(f_n \in L^1\) with \(f_n \to f\) almost everywhere,
        
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          There exist \(g_n\geq 0 \in L^1\) nonnegative with \({\left\lvert {f_n} \right\rvert} \leq g_n\),
        
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          \(g_n\to g\) almost everywhere with \(g\in L^1\), and
        
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          \(\lim \int g_n = \int g\),
        
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      then \(f\in L^1\) and \(\lim \int f_n = \int f < \infty\).
    

  
    

      
    

      Note that this is the DCT with \({\left\lvert {f_n} \right\rvert} < {\left\lvert {g} \right\rvert}\) relaxed to \({\left\lvert {f_n} \right\rvert} < g_n \to g\in L^1\).
    

  
    

  

    

      Proceed by showing \(\limsup \int f_n \leq \int f \leq \liminf \int f_n\):
    

  
    
    
      
        
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      \(\int f \geq \limsup \int f_n\): 

\begin{align*}
\int g - \int f 
&= \int \qty{g-f} \\
&\leq \liminf \int \qty{g_n - f_n} \quad \text{Fatou} \\
&= \lim \int g_n + \liminf \int (-f_n) \\
&= \lim \int g_n - \limsup \int f_n \\
&= \int g - \limsup \int f_n \\
\\
\implies \int f &\geq \limsup \int f_n
.\end{align*}

    
    
  
    
      
      
        
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          Here we use \(g_n - f_n \overset{n\to\infty}\longrightarrow g-f\) with \(0 \leq {\left\lvert {f_n} \right\rvert} - f_n \leq g_n - f_n\), so \(g_n - f_n\) are nonnegative (and measurable) and Fatou’s lemma applies.
        
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      \(\int f \leq \liminf \int f_n\): 

\begin{align*}
\int g + \int f 
&= \int(g+f) \\
&\leq \liminf \int \qty{g_n + f_n} \\
&= \lim \int g_n + \liminf \int f_n \\
&= \int g + \liminf f_n \\
\\
\int f &\leq \liminf \int f_n
.\end{align*}

    
    
  
    
      
      
        
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          Here we use that \(g_n + f_n \to g+f\) with \(0 \leq {\left\lvert {f_n} \right\rvert} + f_n \leq g_n + f_n\) so Fatou’s lemma again applies.
        
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      The converse to the DCT does not hold, i.e. \(L^p\) boundedness does not imply a.e. boundedness, i.e. it is not true that \(\lim \int f_k = \int f\) implies that \(\exists g\in L^p\) such that \(f_k < g\) a.e. for every \(k\).
    

  
    

      Take
    

  
    
    
      
        
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      \(b_k = \sum_{j=1}^k \frac 1 j \to \infty\)
    
    
  
        
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      \(f_k = \chi_{[b_k, b_{k+1}]}\)
    
    
  
        
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      Then
    

  
    
    
      
        
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      \(f_k \overset{a.e.}\to f = 0\),
    
    
  
        
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      \(\int f_k = \frac 1 k \to 0 \implies {\left\lVert {f_k} \right\rVert}_p \to 0\),
    
    
  
        
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      \(0 = \int f = \lim \int f_k = 0\)
    
    
  
        
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      But \(g > f_k \implies g > {\left\lVert {f_k} \right\rVert}_\infty = 1\) a.e. \(\implies g\not\in L^p({\mathbf{R}})\).
    
    
  
        
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      If \(f\in L^1\), then 

\begin{align*}
{\left\lVert {f_n - f} \right\rVert}_{L^1} \overset{n\to\infty}\longrightarrow 0 \iff {\left\lVert {f_n} \right\rVert}_{L^1} \overset{n\to\infty}\longrightarrow {\left\lVert {f} \right\rVert}_{L^1}
.\end{align*}

    

  

    

      Let \(g_n = {\left\lvert {f_n} \right\rvert} - {\left\lvert {f_n - f} \right\rvert}\), then \(g_n \to {\left\lvert {f} \right\rvert}\) and 

\begin{align*}
{\left\lvert {g_n} \right\rvert} = {\left\lvert { {\left\lvert {f_n} \right\rvert} - {\left\lvert {f_n - f} \right\rvert} } \right\rvert} \geq {\left\lvert {f_n - (f_n - f)} \right\rvert} = {\left\lvert {f} \right\rvert} \in L^1
,\end{align*}
 so the DCT applies to \(g_n\) and 

\begin{align*}
{\left\lVert {f_n - f} \right\rVert}_1 = \int {\left\lvert {f_n - f} \right\rvert} + {\left\lvert {f_n} \right\rvert} - {\left\lvert {f_n} \right\rvert}
= \int {\left\lvert {f_n} \right\rvert} - g_n\\
\to_{DCT} \lim \int {\left\lvert {f_n} \right\rvert} - \int {\left\lvert {f} \right\rvert}
.\end{align*}

    

  

    

      If \(f_n\) is a sequence of nonnegative measurable functions, then 

\begin{align*}
\liminf_n \int f_n 
&\geq \int \liminf_n f_n \\
\limsup_n \int f_n &\leq \int \limsup_n f_n
.\end{align*}

    

  

    

      #todo Prove
    

  

    

      

\begin{align*}
\lim _{n \rightarrow \infty} \int_{0}^{\infty} \frac{n^{2}}{1+n^{2} x^{2}} e^{-\frac{x^{2}}{n^{3}}} d x
\overset{\text{Fatou}}\geq
\int_{0}^{\infty} \lim _{n \rightarrow \infty}  \frac{n^{2}}{1+n^{2} x^{2}} e^{-\frac{x^{2}}{n^{3}}} d x \to \int \infty
.\end{align*}

    

  
    

      Note that MCT might work, but showing that this is non-decreasing in \(n\) is difficult.
    

  
Commuting

    

      

\begin{align*}
f_n \geq 0 \text{ and } \sum_n \int {\left\lvert {f_n} \right\rvert} = \sum_n {\left\lVert {f_n} \right\rVert}_{L^1} < \infty \implies \sum_n \int f_n = \int \sum_n f_n
.\end{align*}
 If the \(f_n\) are not necessarily non-negative, we still have 

\begin{align*}  
\left\{{f_n}\right\} \subseteq L^1 \text { and }\qty{\sum\int{\left\lvert {f_n} \right\rvert} < \infty \text { or } \int \sum {\left\lvert {f_n} \right\rvert} < \infty }
\implies
\int\sum_n f_n = \sum_n \int f_n
.\end{align*}

    

  

    
    
      
        
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          Idea: MCT.
        
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          Let \(F_N = \sum^N f_n\) be a finite partial sum;
        
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          Then there are simple functions \(\phi_n \nearrow f_n\)
        
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          So \(\sum^N \phi_n \nearrow F_N\) and MCT applies
        
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          By Tonelli, if \(f_n(x) \geq 0\) for all \(n\), taking the counting measure allows interchanging the order of “integration”.
        
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          By Fubini on \({\left\lvert {f_n} \right\rvert}\), if either “iterated integral” is finite then the result follows.
        
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\begin{align*}
\left\{{f_n}\right\} \subseteq L^1 \text{ and } \sum_n {\left\lVert {f_n} \right\rVert}_{L^1} < \infty \implies \sum_n f_n \text{ converges }{ \text{a.e.} }\text{ and in } L^1
.\end{align*}

    

  

    

      Define \(F_N = \sum^N f_k\) and \(F = \lim_N F_N\), then \({\left\lVert {F_N} \right\rVert}_1 \leq \sum^N {\left\lVert {f_k} \right\rVert} < \infty\) so \(F\in L^1\) and \({\left\lVert {F_N - F} \right\rVert}_1 \to 0\) so the sum converges in \(L^1\). Almost everywhere convergence: ?
    

  

    

      If \(f:X\times I \to {\mathbf{C}}\) where \(f_t: X\to {\mathbf{C}}\) is integrable for each \(t\), then if \({\left\lvert {f(x, t)} \right\rvert} \leq {\left\lvert {g(x)} \right\rvert}\) for some \(g\in L^1\), then 

\begin{align*}
\lim_{t\to t_0}\int_X f(x, t) \,d\mu= \int_X f(x, t_0) \,d\mu\coloneqq F(t_0)
,\end{align*}
 and if \(f_x: I\to {\mathbf{C}}\) is continuous for all \(x\), then \(F: I\to {\mathbf{C}}\) is continuous.
    

  
    

      Moreover if \({\frac{\partial f}{\partial t}\,}\) exists and \({\left\lvert {{\frac{\partial f}{\partial t}\,}(x, t)} \right\rvert} \leq {\left\lvert {g} \right\rvert}\) for some \(g\in L^1\), then 

\begin{align*}
{\frac{\partial }{\partial t}\,} \int_X f(x, t) \,d\mu
= \int_X {\frac{\partial }{\partial t}\,} f(x, t) \,d\mu
.\end{align*}

    

  

  


            

              
              
            

            
  

    
      
        
        #todo