Let ( X , σ , μ ) {\displaystyle (X,\sigma ,\mu )} be a σ {\displaystyle \sigma } -finite measure space. Suppose s {\displaystyle s} is a positive simple measurable function, with s = ∑ i = 1 3 y i χ A i {\displaystyle s=\displaystyle \sum _{i=1}^{3}y_{i}\chi _{A_{i}}} ; A i ∈ σ {\displaystyle A_{i}\in \sigma } are disjoint.
Define ∫ X s d μ = ∑ y i μ ( A i ) {\displaystyle \displaystyle \int _{X}s~d\mu =\sum y_{i}\mu (A_{i})}
Let f : X → R ¯ {\displaystyle f:X\to {\overline {\mathbb {R} }}} be measurable, and let f ≥ 0 {\displaystyle f\geq 0} .
Define ∫ X f d μ = sup { ∫ X s d μ = ∑ y i μ ( A i ) | s simple , s ≥ 0 , s ≤ f } {\displaystyle \displaystyle \int _{X}f~d\mu =\sup\{\int _{X}s~d\mu =\sum y_{i}\mu (A_{i})|s{\text{ simple }},s\geq 0,s\leq f\}}
Now let f {\displaystyle f} be any measurable function. We say that f {\displaystyle f} is integrable if f + {\displaystyle f^{+}} and f − {\displaystyle f^{-}} are integrable and if ∫ X f + d μ , ∫ X f − d μ < ∞ {\displaystyle \displaystyle \int _{X}f^{+}~d\mu ,\int _{X}f^{-}~d\mu <\infty } . Then, we write
∫ X f d μ = ∫ X f + d μ − ∫ X f − d μ {\displaystyle \displaystyle \int _{X}f~d\mu =\int _{X}f^{+}~d\mu -\int _{X}f^{-}~d\mu }
The class of measurable functions on X {\displaystyle X} is denoted by L 1 ( X ) {\displaystyle {\mathcal {L}}^{1}(X)}
For 0 < p < ∞ {\displaystyle 0<p<\infty } , we define L p {\displaystyle {\mathcal {L}}^{p}} to be the collection of all measurable functions f {\displaystyle f} such that | f | p ∈ L 1 {\displaystyle |f|^{p}\in {\mathcal {L}}^{1}}
A property is said to hold almost everywhere if the set of all points where the property does not hold has measure zero.
Properties
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Let ( X , σ , μ ) {\displaystyle (X,\sigma ,\mu )} be a measure space and let f , g {\displaystyle f,g} be measurable on X {\displaystyle X} . Then
If f ≤ g {\displaystyle f\leq g} , then ∫ X f d μ ≤ ∫ X g d μ {\displaystyle \displaystyle \int _{X}fd\mu \leq \int _{X}gd\mu }
If A , B ∈ σ {\displaystyle A,B\in \sigma } , A ⊂ B {\displaystyle A\subset B} , then ∫ A f d μ ≤ ∫ B f d μ {\displaystyle \displaystyle \int _{A}fd\mu \leq \int _{B}fd\mu }
If f ≥ 0 {\displaystyle f\geq 0} and c ≥ 0 {\displaystyle c\geq 0} then ∫ X c f d μ = c ∫ X f d μ {\displaystyle \displaystyle \int _{X}cfd\mu =c\int _{X}fd\mu }
If E ∈ σ {\displaystyle E\in \sigma } , μ ( E ) = 0 {\displaystyle \mu (E)=0} , then ∫ E f d μ = 0 {\displaystyle \displaystyle \int _{E}fd\mu =0} , even if f ( E ) = { ∞ } {\displaystyle f(E)=\{\infty \}}
If E ∈ σ {\displaystyle E\in \sigma } , f ( E ) = { 0 } {\displaystyle f(E)=\{0\}} , then ∫ E f d μ = 0 {\displaystyle \displaystyle \int _{E}fd\mu =0} , even if μ ( E ) = ∞ {\displaystyle \mu (E)=\infty } Proof
Monotone Convergence Theorem
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Suppose f n ≥ 0 {\displaystyle f_{n}\geq 0} and f n {\displaystyle f_{n}} are measurable for all n {\displaystyle n} such that
f 1 ( x ) ≤ f 2 ( x ) ≤ … {\displaystyle f_{1}(x)\leq f_{2}(x)\leq \ldots } for every x ∈ X {\displaystyle x\in X}
f n ( x ) → f ( x ) {\displaystyle f_{n}(x)\to f(x)} almost everywhere on X {\displaystyle X} Then, ∫ X f n d μ → ∫ X f d μ {\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \int _{X}fd\mu }
Proof
∫ X f n d μ {\displaystyle \displaystyle \int _{X}f_{n}d\mu } is an increasing sequence in R {\displaystyle \mathbb {R} } , and hence, ∫ X f n d μ → α ∈ R ¯ {\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \alpha \in {\overline {\mathbb {R} }}} (say). We know that f {\displaystyle f} is measurable and that f ≥ f n ∀ n {\displaystyle f\geq f_{n}\forall n} . That is,
∫ X f 1 d μ ≤ ∫ X f 2 d μ ≤ … ∫ X f n d μ ≤ … ∫ X f d μ {\displaystyle \displaystyle \int _{X}f_{1}d\mu \leq \int _{X}f_{2}d\mu \leq \ldots \int _{X}f_{n}d\mu \leq \ldots \int _{X}fd\mu }
Hence, α ≤ ∫ X f d μ = sup { ∫ X s d μ : s is simple , 0 ≤ s ≤ 1 } {\displaystyle \displaystyle \alpha \leq \int _{X}fd\mu =\sup\{\int _{X}sd\mu :s{\text{ is simple }},0\leq s\leq 1\}}
Let c ∈ [ 0 , 1 ] {\displaystyle c\in [0,1]}
Define E n = { x | f n ( x ) ≥ c s ( x ) } {\displaystyle E_{n}=\{x|f_{n}(x)\geq cs(x)\}} ; n = 1 , 2 … {\displaystyle n=1,2\ldots } . Observe that E 1 ⊂ E 2 ⊂ … {\displaystyle E_{1}\subset E_{2}\subset \ldots } and ⋃ E n = X {\displaystyle \bigcup E_{n}=X}
Suppose x ∈ X {\displaystyle x\in X} . If f ( x ) = 0 {\displaystyle f(x)=0} then s ( x ) = 0 {\displaystyle s(x)=0} implying that x ∈ E 1 {\displaystyle x\in E_{1}} . If f ( x ) > 0 {\displaystyle f(x)>0} , then there exists n {\displaystyle n} such that f n ( x ) > c s ( x ) {\displaystyle f_{n}(x)>cs(x)} and hence, x ∈ E n {\displaystyle x\in E_{n}} .
Thus, ⋃ E n = X {\displaystyle \bigcup E_{n}=X} , therefore ∫ X f n ( x ) d μ ≥ ∫ E n f n d μ ≥ c ∫ E n s d μ {\displaystyle \displaystyle \int _{X}f_{n}(x)d\mu \geq \int _{E_{n}}f_{n}d\mu \geq c\int _{E_{n}}sd\mu } . As this is true if c ∈ [ 0 , 1 ] {\displaystyle c\in [0,1]} , we have that α ≥ ∫ X f d μ {\displaystyle \alpha \geq \displaystyle \int _{X}fd\mu } . Thus, ∫ X f n d μ → ∫ X f d μ {\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \int _{X}fd\mu } .
Fatou's Lemma
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Dominated convergence theorem
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Let ( X , σ , μ ) {\displaystyle (X,\sigma ,\mu )} be a complex measure space. Let { f n } {\displaystyle \{f_{n}\}} be a sequence of complex measurable functions that converge pointwise to f {\displaystyle f} ; f ( x ) = lim n → ∞ f n ( x ) {\displaystyle f(x)=\displaystyle \lim _{n\to \infty }f_{n}(x)} , with x ∈ X {\displaystyle x\in X}
Suppose | f n ( x ) | ≤ g ( x ) {\displaystyle |f_{n}(x)|\leq g(x)} for some g ∈ L 1 ( X ) {\displaystyle g\in {\mathcal {L}}^{1}(X)} then
f ∈ L 1 {\displaystyle f\in {\mathcal {L}}^{1}} and ∫ X | f n − f | d μ → 0 {\displaystyle \displaystyle \int _{X}|f_{n}-f|d\mu \to 0} as n → ∞ {\displaystyle n\to \infty }
Proof
We know that | f | ≤ g {\displaystyle |f|\leq g} and hence | f n − f | ≤ 2 g {\displaystyle |f_{n}-f|\leq 2g} , that is, 0 ≤ 2 g − | f n − f | {\displaystyle 0\leq 2g-|f_{n}-f|}
Therefore, by Fatou's lemma, ∫ X 2 g d μ ≤ lim inf ∫ X ( 2 g − | f n − f | ) d μ ≤ ∫ X 2 g d μ + lim inf ∫ X ( − | f n − f | ) d μ {\displaystyle \displaystyle \int _{X}2gd\mu \leq \liminf \int _{X}(2g-|f_{n}-f|)d\mu \leq \displaystyle \int _{X}2gd\mu +\liminf \int _{X}(-|f_{n}-f|)d\mu }
= ∫ X 2 g d μ − lim sup ∫ X | f n − f | d μ {\displaystyle =\displaystyle \int _{X}2gd\mu -\limsup \int _{X}|f_{n}-f|d\mu }
As g ∈ L 1 {\displaystyle g\in {\mathcal {L}}^{1}} , lim sup n → ∞ ∫ X | f n − f | d μ ≤ {\displaystyle \displaystyle \limsup _{n\to \infty }\int _{X}|f_{n}-f|d\mu \leq } implying that lim sup ∫ X | f n − f | d μ {\displaystyle \displaystyle \limsup \int _{X}|f_{n}-f|d\mu }
Suppose f : X → [ 0 , ∞ ] {\displaystyle f:X\to [0,\infty ]} is measurable, E ∈ σ {\displaystyle E\in \sigma } with μ ( E ) > 0 {\displaystyle \mu (E)>0} such that ∫ E f d μ = 0 {\displaystyle \displaystyle \int _{E}fd\mu =0} . Then f ( x ) = 0 {\displaystyle f(x)=0} almost everywhere E {\displaystyle E}
Let f ∈ L 1 ( X ) {\displaystyle f\in {\mathcal {L}}^{1}(X)} and let ∫ E f d μ = 0 {\displaystyle \displaystyle \int _{E}fd\mu =0} for every E ∈ σ {\displaystyle E\in \sigma } . Then, f = 0 {\displaystyle f=0} almost everywhere on X {\displaystyle X}
Let f ∈ L 1 ( X ) {\displaystyle f\in {\mathcal {L}}^{1}(X)} and | ∫ X f d μ | = ∫ X | f | d μ {\displaystyle \displaystyle \left|\int _{X}fd\mu \right|=\int _{X}|f|d\mu } then there exists constant α {\displaystyle \alpha } such that | f | = α f {\displaystyle |f|=\alpha f} almost everywhere on E {\displaystyle E} Proof
For each n ∈ N {\displaystyle n\in \mathbb {N} } define A n = { x ∈ E | f ( x ) > 1 n } {\displaystyle A_{n}=\{x\in E|f(x)>{\frac {1}{n}}\}} . Observe that A n ↑ E {\displaystyle A_{n}\uparrow E} but 1 n μ ( A n ) ≤ ∫ A n f d μ ≤ ∫ E f d μ = 0 {\displaystyle {\frac {1}{n}}\mu (A_{n})\leq \displaystyle \int _{A_{n}}fd\mu \leq \int _{E}fd\mu =0} Thus μ ( A n ) = 0 {\displaystyle \mu (A_{n})=0} for all n {\displaystyle n} , by continuity, f = 0 {\displaystyle f=0} almost everywhere on E {\displaystyle E}
Write ∫ E f d μ = ∫ E u + d μ − ∫ E u − d μ + i ( ∫ E v + d μ − ∫ E v − d μ ) {\displaystyle \displaystyle \int _{E}fd\mu =\int _{E}u^{+}d\mu -\int _{E}u^{-}d\mu +i\left(\int _{E}v^{+}d\mu -\int _{E}v^{-}d\mu \right)} , where u + , u − , v + , v − {\displaystyle u^{+},u^{-},v^{+},v^{-}} are non-negative real measurable. Further as ∫ E u + d μ , ∫ E ( − u − ) d μ {\displaystyle \displaystyle \int _{E}u^{+}d\mu ,\int _{E}(-u^{-})d\mu } are both non-negative, each of them is zero. Thus, by applying part I, we have that u + , u − {\displaystyle u^{+},u^{-}} vanish almost everywhere on E {\displaystyle E} . We can similarly show that v + , v − {\displaystyle v^{+},v^{-}} vanish almost everywhere on E {\displaystyle E} .