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Lebesgue Integral

Definition (Integral of Simple Functions)

Let (X,F,μ)(X,\mathcal{F},\mu) be a Measure Space, let E∈FE\in\mathcal{F}. Let f:X→R+f:X\to \mathbb{R}^{+} be a Simple Function: f=∑i=1nai1Aif = \sum_{i=1}^{n}a_{i}\mathbb{1}_{A_{i}} where Ai=f−1({αi}), ∀1≤i≤nA_{i}=f^{-1}(\{ \alpha_{i} \}),\ \forall_{1}\le i\le n. The Lebesgue Integral of ff with respect to measure μ\mu is defined as: ∫Ef dμ=∑i=1nai μ(Ai∩E)\int\limits _{E}f \, d\mu =\sum_{i=1}^{n}a_{i} \ \mu(A_{i}\cap E)where Ai∈F,ai≥0A_{i}\in\mathcal{F},a_{i}\ge0.

Lemma (Monotonicity for integrals of simple functions)

Let S1,S2S_{1},S_{2} be two Simple Functions s.t. S1(x)≤S2(x), ∀x∈XS_{1}(x)\le S_{2}(x),\ \forall x \in X. Then ∫S1 dμ≤∫S2 dμ\int\limits S_{1} \, d\mu\le \int\limits S_{2} \, d\mu

Definition (Integral of Positive Measurable Functions)

We can further the definition of where for f:X→[0,∞)f:X\to[0,\infty) measurable we define the Lebesgue Integral as: ∫Xf dμ=sup⁡{∫Xg dμ:0≤g<f, g simple}\int\limits _{X}f \, d\mu =\sup\left\{ \int\limits _{X}g \, d\mu :0\le g<f, \ g \ \text{simple} \right\}

Definition (Lebesgue integral)

Finally, adding on to the Integral of Positive Measurable Functions, let f:X→Rf:X\to \mathbb{R} be measurable. Let f+=max(f,0)f^{+}=max(f,0), f−=max(−f,0)f^{-}=max(-f,0) we define the Lebesgue Integral as: ∫Xf dμ=∫Xf+ dμ−∫Xf− dμ\int\limits _{X}f \, d\mu=\int\limits _{X}f^+ \, d\mu-\int\limits _{X}f^- \, d\mu

Theorem (1.24)

Let (X,M,μ)(X,\mathcal{M},\mu) be a Measure Space. Let f,gf,g be measurable functions. Then,

  1. Monotonicity: If 0≤f≤g0\le f\le g, then ∫Ef dμ≤∫Eg dμ\int\limits _{E}f \, d\mu \le\int\limits _{E}g \, d\mu
  2. Monotonicity: If A⊆BA\subseteq B, and f≥0f\ge 0 then, ∫Af dμ≤∫Bf dμ\int\limits _{A}f \, d\mu\le\int\limits _{B}f \, d\mu
  3. Linearity: If f≥0f\ge 0 and cc is a constant, 0≤c<∞0\le c<\infty then c∫f dμ=∫cf dμc\int\limits f \, d\mu=\int\limits cf \, d\mu
  4. If f(x)=0, ∀x∈Ef(x)=0,\ \forall x\in E, then ∫Ef dμ=0\int\limits _{E}f \, d\mu=0even if μ(E)=∞\mu(E)=\infty.
  5. If μ(E)=0\mu(E)=0 then ∫Ef dμ=0\int\limits _{E}f \, d\mu=0 even if f(x)=∞f(x)=\infty, ∀x∈E\forall x \in E.
  6. ∫Ef dμ=∫Xf⋅1E dμ\int\limits _{E}f \, d\mu=\int\limits_{X} f\cdot\mathbb{1}_{E} \, d\mu

Proposition (1.25)

Let s,ts,t be nonnegative measurable simple functions on XX. For E∈ME\in\mathcal{M}, define ν(E)=∫Es dμ\nu(E)=\int\limits _{E}s \, d\mu Then ν\nu is a Measure on M\mathcal{M}. Also ∫X(s+t) dμ=∫Xs dμ+∫Xt dμ\int\limits _{X}(s+t) \, d\mu=\int\limits _{X}s \, d\mu+\int\limits _{X}t \, d\mu

Theorem (1.29)

Suppose f:X→[0,+∞]f:X\to[0,+\infty] is measurable, and ν(E)=∫Ef dμ(E∈M)\nu(E)=\int\limits _{E}f \, d\mu \quad (E \in \mathcal{M})Then ν\nu is a Measure on M\mathscr{M} and ∫Xg dν=∫Xg⋅f dμ\int\limits _{X}g \, d\nu =\int\limits _{X}g\cdot f \, d\mu for every measurable gg on XX with range in [0,+∞][0,+\infty].

Remark

  • The converse of this theorem is the Radon-Nikodym Theorem and as a result is intimately connected.
  • We call ff the density of ν\nu w.r.t. μ\mu and by the second identity we can write dν=f dμd\nu=f\,d\muordνdμ=f\frac{d\nu}{d\mu}=fwhich is commonly referred to as the Radon-Nikodym Derivative

Definition (Complex Lebesgue Integral)

If f:X→Cf:X\to \mathbb{C} is s.t. f=u+ivf=u+iv and u,v:X→Ru,v:X\to \mathbb{R} are Measurable Functions, and if f∈L1(X,M,μ)f\in\mathscr{L}^{1}(X,\mathscr{M},\mu) then ∫Ef dμ=∫Eu+ dμ−∫Eu− dμ+i∫Ev+ dμ−i∫Ev− dμ\int\limits _{E}f \, d\mu=\int\limits _{E}u^{+} \, d\mu -\int\limits _{E}u^{-} \, d\mu+i\int\limits_{E}v^{+}\,d\mu-i \int\limits _{E}v^{-} \, d\mu for all E∈ME\in\mathscr{M}.

Note:

f measurable  ⟹  f+,f− measurablef\text{ measurable}\implies f^{+},f^{-}\text{ measurable}by Composition of Measurable Functions. This is only possible with Borel σ-algebra hence why we restrict ourselves to this instead of Lebesgue Measurable. >[!prp] Properties of Lebesgue integral >1. Linearity: I(αf+βg)=αI(f)+βI(g)I(\alpha f+\beta g)=\alpha I(f)+\beta I(g) >2. Monotonicity: f≤g  ⟹  I(f)≤I(g)f\le g\implies I(f)\le I(g)

Theorem (Absolute Continuity of Lebesgue Integral)

Let (X,F,μ)(X,\mathcal{F},\mu) be a measure space. Let f∈L1(X,F,μ)f\in\mathscr{L}^{1}(X,\mathcal{F},\mu). Then ∀ϵ>0 ∃δ>0 s.t. ∀A∈F\forall\epsilon>0 \ \exists\delta>0\text{ s.t. }\forall A\in\mathcal{F} μ(A)≤δ  ⟹  ∫A∣f∣ dμ≤ϵ\mu(A)\le\delta \implies \int\limits _{A}|f| \, d\mu\le\epsilon i.e. The Lebesgue Integral is absolutely continuous w.r.t. the Lebesgue Measure ∫ dμ≪μ\int\limits \, d\mu\ll\mu

Theorem (1.33)

If f∈L1(X,M,μ)f\in\mathscr{L}^{1}(X,\mathscr{M},\mu) then ∣∫Xf dμ∣≤∫X∣f∣ dμ\left|\int\limits _{X}f \, d\mu \right|\le \int\limits _{X}|f| \, d\mu

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