Corollary 4.3. (Baby MCT) Let \(f\) be a non-negative measurable function, \((E_n)\) be an increasing sequence of measurable sets, and \(E=\bigcup_{i=1}^∞ E_n\). Then \(\int_E f=\lim_{n→∞} \int_{E_n} f=\sup_n \int_{E_n} f\).

Proof. Apply MCT to \((f χ_{E_n})\). ◻

Corollary 4.4. (Additivity) For \(0⩽f, g\) measurable, \(\int (f+g)=\int f+\int g\).

Proof. By Prop 3.9 there exists simple functions \(0⩽ϕ_1⩽ϕ_2⩽⋯↗f\) and \(0⩽ψ_1⩽ψ_2⩽⋯↗g\)

i.e. \(ϕ_n (x)↗f (x)\) and \(ψ_n (x)↗g (x)\) for all \(x\).

Corollary 4.5. (MCT for series) Let \((f_n)^∞_{n=1}\) be non-negative measurable functions, \(f=\sum_{n=1}^∞ f_n\). Then \(\int f=\sum_{n=1}^∞ f_n\). [\(f\) is integrable\(⇔\sum_{n=1}^∞ f_n\) converges]

Corollary 4.6. Let \(f : [a, b]→[0,∞)\) be continuous. Then \(\int_{[a, b]} f=\int_a^b f\) is equal to the R-integral of \(f\) on \([a, b]\).

Proof. Sucessively bisecting \([a, b]\). Let \(ϕ_n\) be the step function corresponding to the lower Riemann sum for the \(n\)-th partition. Then \(0⩽ϕ_1⩽ϕ_2⩽⋯↗f\). The R-integral is equal to \(\lim_{n→∞} \int_a^b ϕ_n\) which is \(\int_a^b f\) by MCT. ◻

Example 4.7. \(f (x)=(1-x)^{-1/2}\) on \([0, 1)\). \(f\) is continuous on \([0, 1)\) so measurable. By Baby MCT \(\int_0^1 (1-x)^{-1/2} \mathrm{d} x=\lim_{n→∞} \int_0^{1-\frac{1}{n}} (1-x)^{-1/2} \mathrm{d} x=\lim_{n→∞} 2 (1-n^{-1/2})=2\) by FTC.

On the other hand, \((1-x)^{-1/2}=\sum_{n=0}^∞ \frac{(2 n) !}{4^n (n!)^2} x^n\) for \(0⩽x<1\)

By MCT for series, \(2=\int_0^1 (1-x)^{-1/2} \mathrm{d} x=\sum_{n=0}^∞ \frac{(2 n) !}{4^n (n!)^2} \int_0^1 x^n \mathrm{d} x=\sum_{n=0}^∞ \frac{(2 n) !}{4^n n! (n+1) !}\)

The fact that the series above converges to 2 can be obtained directly from the Binomial Expansion of \((1-x)^{1/2}\), via Abel's theorem.

Example 4.8. \(\lim_{n→∞} \int_0^{nπ} \cos \left( \frac{x}{2 n} \right) x^2 \mathrm{e}^{-x^3} \mathrm{d} x\)

Let \(f_n (x)=\cos \left( \frac{x}{2 n} \right) x^2 \mathrm{e}^{-x^3} χ_{[0, nπ]} (x)\)

These are measurable as \(\cos \left( \frac{x}{2 n} \right) x^2 \mathrm{e}^{-x^3}\) is continuous.

\(0⩽f_1⩽f_2⩽⋯\) as \(\cos \left( \frac{x}{2 n} \right)⩽\cos \left( \frac{x}{2 (n+1)} \right)\) for \(0⩽x⩽nπ\)

By MCT \(\lim_{n→∞} \int_0^{nπ} \cos \left( \frac{x}{2 n} \right) x^2 \mathrm{e}^{-x^3} \mathrm{d} x=\int_0^∞ x^2 \mathrm{e}^{-x^3} \mathrm{d} x\) as \(f_n (x)→x^2 \mathrm{e}^{-x^3}\) pointwise.

Now \(\int_0^∞ x^2 \mathrm{e}^{-x^3} \mathrm{d} x=\lim_{n→∞}\int_0^n x^2 \mathrm{e}^{-x^3} \mathrm{d} x\) by Baby MCT

\(=\lim_{n→∞}\frac{1-\mathrm e^{- n^3}}{3}=\frac{1}{3}\) by FTC.

5 Lebesgue Integral: General functions

Definition. For \(f : ℝ→[-∞,∞]\) measurable, set \(f^+=\max (f, 0), f^-=\max (- f, 0)\) both are non-negative and measurable. Note \({|f|}=f^++f^-\) and \(f=f^+-f^-\). Say \(f\) is measurable over \(ℝ\), written as \(f =ℒ^1(ℝ)\). If \(f\) is measurable and \(\int f^+, \int f^-\) are finite. In this case define \(\int f=\int f^+-\int f^-\).

Say \(E∈ℳ_\text{Leb}\), \(f\) is integrable over \(E\), write \(f∈ℒ^1(E)\), when \(f χ_E∈ℒ^1(ℝ)\). In this case \(\int_E f=\int_ℝf χ_E\).

Proposition 5.1.

1. If \(f\) is integrable, then \(|f|\) is integrable. [drawback of Lebesgue theory]
2. \({|f|}∈ℒ^1(ℝ)\) and \(f\) is measurable\(⇒f∈ℒ^1(ℝ)\).
3. (Comparison Test) Let \(f : ℝ→[-∞,∞]\) be measurable,
   • If \({|f|}⩽g\), for some \(g∈ℒ^1(ℝ)\), then \(f∈ℒ^1(ℝ)\).
   • Suppose \({|f|}⩾g⩾0\) for some measurable \(g\) which is not integrable, then \(f\) is not integrable.
4. If \(f, g∈ℒ^1(ℝ)\) and \(f+g\) is defined. Then \(f+g∈ℒ^1(ℝ)\) and \(αf∈ℒ^1(ℝ)\) for \(α∈ℝ\). Moreover, \(∫(f+g)=∫f+∫g,∫αf=α∫f\).
5. If \(f∈ℒ^1(ℝ)\) and \(g=f\) a.e. then \(g∈ℒ^1(ℝ)\) and \(∫f=∫g\).
6. If \(f\) is integrable\(⇒f (x)∈ℝ\) a.e.
7. If \(f\) is integrable \(∫{|f|}=0⇒f=0\) a.e.
8. If \(f\) is integrable over \(E∈ℳ_\text{Leb}\) and \(E_1⊆E_2⊆E_3⊆…\) are measurable with \(\bigcup_{n=1}^∞ E_n=E\). Then \(∫_E f=\lim_{n→∞}∫_{E_n} f\).