Third step let ℝ and let max 0 and min 0 if is not

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Third step Let ? ∶ (?, 𝒜) → ℝ and ? = ? + − ? . Let ? + = max {?, 0} and ? = min {?, 0} . If ? ? + ?? = ∫ ? ? ?? = ∞ ∶ ∫ ? ? ?? is not defined. Otherwise ? ? ?? = ∫ ? ? + ?? − ∫ ? ? ?? . Does this definition/construction of the Lebesgue integral satisfy the desired proper- ties of linearity/monotonicity/…? In the following, we will denote “simple” functions always as ? . Definition 3.2. Let ? ∶ (?, 𝒜) → [0, ∞] be measurable. Let ? ∈ 𝒜 . ? ? ?? B 1 ? ? ?? Lemma 3.3. Let ? ∶ (?, 𝒜) → [0, ∞] be a simple function. Then ? ? (?) = ∫ ? ? ?? is a measure on (?, 𝒜) . ? ? (?) = ? ?=1 ? ? ?([? = ? ? ] ∩ ?) because 1 ? ⋅ ? = ∑ ? ?=1 ? ? 1 [?=? ? ] 1 ? + 0 ⋅ 1 ? 𝐶 . ? ↦ ?([? = ? ? ] ∩ ?) is a measure ∀? . This lecture took place on 2018/10/22. Definition 3.4. Let (?, 𝒜, ?) be a measure space. ? ∶ (?, 𝒜) → (ℝ, ℬ) is called simple if ?(?) is finite. ? ≥ 0 . ? ? ?? B ? ? ⋅ ?[? = ?] Trivial: If ? = ∑ ? ?=1 ? ? 1 ? ? , ? ? ∈ 𝒜 then ? is simple. ? ? are not necessarily pairwise disjoint and ? ? ?? = ∑ ? ?=1 ? ? ?(? ? ) . Proof. 𝜀 ∈ {−1, 1} ? with ? 1 B ?, ? −1 B ? ? . 𝜀 = (𝜀 1 , … , 𝜀 ? ) . E.g. ? 1 ∩ ? 2 ? ? 3 = ? 1,1,−1 . ? 𝜀 = ? 𝜀 1 1 ∩ ? 𝜀 2 2 ∩ ⋯ ∩ ? 𝜀 𝑚 ? 18
is pairwise disjoint. On ? 𝜀 , ? has value 𝜀 ? =1 ? ? = ? 𝜀 ⟹ ? = ∑ ? 𝜀 1 ? ⃗𝜀 and ∫ ≤ ?? = ∑ 𝜀 ? 𝜀 ?(? 𝜀 ) = ⋯ = ∑ ? ? ?(? ? ) (where 𝜀 ? 𝜀 ?(? 𝜀 ) is the disjoint case and ∑ ? ? ?(? ? ) is generic). 𝜀 ?∶𝜀 ? =1 ? ? ⋅ ?(? 𝜀 ) = ∑ ? ? ? 𝜀∶𝜀 ? =1 ?(? 𝜀 ) = ∑ ? ? ? ?(? ? ) Corollary 3.5. Let ? 1 , ? 2 ∶ ? → [0, ∞] be simple. Then ? = 𝛼 ⋅ ? 1 + 𝛽 ⋅ ? 2 ( 𝛼, 𝛽 ≥ 0 ) is simple and ∫ ? ?? = 𝛼 ⋅ ∫ ? 1 ?? + 𝛽 ∫ ? 2 ?? . Theorem 3.5.1 (Markov inequality) . Let ? ∈ ℝ . Let ? ≥ 0 . ? ⋅ ?[ ? ≥ ? {?∈? ∣ 𝑓(?)≥?} ] ≤ ∫ ? ?? Proof. ? = ? ⋅ 1 [𝑓≥?] ≤ ? If ? ∈ [? ≤ ?] ∶ ? ⋅ 1 ≤ ?(?) . If ? ∉ [? ≤ ?] ∶ ? ⋅ 0 ≤ ?(?) . ? is simple, so ??[? ≥ ?] = ∫ ? ?? ≤ ∫ ? ?? . ? = 0 ∶ 1 [𝑓<?] × ? ⋅ 1 [𝑓≥?] . Definition 3.6. A statement holds almost everywhere if ∀? ∈ 𝒜 ∶ ?(? ? ) = 0 . So ? ? is a null set, i.e. of measure zero. Theorem 3.6.1. ∀?, ? ∶ ? → [0, ∞] measurable ? ≤ ? almost everywhere ⟹ ∫ ? ?? ≤ ∫ ? ?? 1. 2. ? = ? almost everywhere ⟹ ∫ ? ?? = ∫ ? ?? 3. ∫ ? ?? = 0 ⟹ ? = 0 almost everywhere 4. ∫ ? ?? < ∞ ⟹ ? < ∞ almost everywhere Proof. 1. Let ? be simple, 0 ≤ ? ≤ ? . ? ⋅ 1 [𝑓≤𝑔] ≤ ? where ? ⋅ 1 [𝑓≤𝑔] is simple. ∫ ? ⋅ 1 [𝑓≤𝑔] ?? ≤ ∫ ? ?? . ∫ ? ⋅ 1 [𝑓≤𝑔] ?? = ∫ ? ?? . If ∀? simple, 0 ≤ ? ≤ ? , then ∫ ? ?? = sup {∫ ? ?? ∣ 0 ≤ ? ≤ ?, ? simple } ≤ ∫ ? ?? 2. ? ≤ ? almost everywhere and ? ≥ ? almost everywhere ⟹ ∫ ? ?? = ∫ ? ?? . 19
3. Markov inequality with ? = 1 ? . 1 𝑛 ? [? ≥ 1 𝑛 ] ≤ ∫ ? ?? = 0 ⟹ ? [? ≥ 1 𝑛 ] = 0∀𝑛 ∈ ℕ ? ∈ [? ≥ 1 ? ] ⟹ ? ∈ [? ≥ 1 ?+1 ] ⟹ ? [? ≥ 1 𝑛 ] → ? [⋃ [? ≥ 1 𝑛 ]] = ? [? > 0] = 0 4. ? > 0 , ? = ? ⋅ 1 [𝑓=∞] ≤ ? . ??[? = ∞] = ∫ ? ?? ≤ ∫ ? ?? = ? < ∞ ?[? = ∞] ≤ ? ? ∀? > 0 ⟹ ?[? = ∞] = 0 Theorem 3.6.2 (Levi’s theorem about monotone convergence) . If ? ? ∶ (?, 𝒜) → [0, ∞] is measurable and pointwise monotonically increasing ( ? 1 ≤ ? 2 ≤ … ) and ? = lim ?→∞ ? ? then ∫ ? ?? = lim ?→∞ ∫ ? ? ?? Proof. Because of (1) in the previous theorem, ∫ ? ? ?? is monotonically increasing and ≤ ∫ ? ?? , so lim ∫ ? ? ?? ≤ ∫ ? ?? . (?) + denotes the function ? if ? ≥ 0 and 0 otherwise. Show “ ”. Let 0 ≤ ? ≤ ? be simple. Let 𝜀 > 0 . ? ?,𝜀 B (? − 𝜀) + 1 [𝑓 𝑛 ≥𝑓−𝜀] is a simple function. ? − 𝜀 ≤ ? − 𝜀 ≤ ? ? . ? ?,𝜀 ≤ ? ? . ? (? − 𝜀) + ? [? = ?, ? ? > ? − 𝜀] = ∫ ? ?,𝜀 ?? ≤ ∫ ? ? ?? ≤ lim ∫ ? ? ?? ? ?,𝜀 = ? (values of s) (? − 𝜀) + 1 [?−?] ⏟⏟⏟⏟⏟⏟⏟⏟⏟ (?−𝜀) + 1 [𝑓 𝑛 >𝑓−𝜀] [? ? > ? − 𝜀] ↗ ? [? = ?, ? ? > ? − 𝜀] ↗ [? − ?] ⟹ ∑ ? (? − 𝜀) + ?[? = ?] ≤ lim ∫ ? ? ?? 𝜀 → 0 ⟹ ∑ ? ??[? = ?] ≤ lim ∫ ? ? ?? If ? > 0 , such that ?[? = ?] = +∞ . 0 < 𝜀 < ? . Let ? ?,𝜀 = (? − 𝜀) + 1 [𝑓 𝑛 ≥?∧(𝑓−𝜀)] , where ? ∧ ? denotes the minimum of ? and ? . Let ? ≥ max ? . 20
This lecture took place on 2018/10/29. Remark (Revision) . Let ? be a simple function. ? = ∑ ? ?=1 ? ? 1 ? ? . ? = ∑ ? ? 1 [?=?] is a finite sum ∫ ? ?? = ∑ ? ?[? = ?] = ∑ ? ?=1 ? ? ?(? ? ) This is independent of the representation. Let ? ∶ (?, 𝒜) → [0, ∞] be measurable. Then we can approximate the integral of ? using the integrals of simple functions. ∫ ? ?? = sup {∫ ? ?? ∣ 0 ≤ ? ≤ ?, simple } Remark (Properties) . 1. 0 ≤ ? ≤ ? almost everywhere (wrt. ? ) ∫ ? ?? ≤ ∫ ? ?? 2. ? = ? almost everywhere (wrt. ? ) ⟹ ∫ ? ?? = ∫ ? ?? 3. ∫ ? ?? = 0 ⟺ ? = 0 almost everywhere (wrt. ? ) 4. ∫ ? ?? < ∞ ⟹ ? < ∞ almost everywhere It is obvious if ? is simple, then ∫ ? ?? = max {∫ ? ?? ∣ 0 ≤ ? ≤ ? simple } Theorem (Monotonic convergence) . Let ?

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