# A (terse) introduction to Lebesgue integration by John Franks

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Extra resources for A (terse) introduction to Lebesgue integration

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If f (x) = fact that ri XAi , the absolute value property follows from the f dµ = ri µ(Ai ) ≤ |ri |µ(Ai ) = |f | dµ. We would like to consider the measure µ(A) for any set A ∈ M, not just subsets of I. Mostly this is straightforward, but there is one 44 3. The Lebesgue Integral notational issue. Some subsets of R have inﬁnite measure, for example, the open interval (0, ∞) or R itself. Hence, we cannot describe µ as a real-valued function deﬁned for any set A ∈ M, because µ(A) might be inﬁnite. There are other instances also when we want to allow the value of a function to be +∞ or −∞.

2) h |h| But the continuity of f implies that given x0 and any ε > 0 there exists δ > 0 such that whenever |t−x0 | < δ we have |f (t)−f (x0 )| < ε. Thus, if |h| < δ, then |f (t)−f (x0 )| < ε for all t between x0 and x0 +h. 6. The Fundamental Theorem of Calculus 15 and hence that x0 +h x0 |f (t) − f (x0 )| dt |h| < ε. 2) above we have that F (x0 + h) − F (x0 ) − f (x0 ) < ε h whenever |h| < δ, which is exactly what we needed to show. 2. (Fundamental theorem of calculus). If f is a continuous function on [a, b] and F is any anti-derivative of f, then b a f (x) dx = F (b) − F (a).

In other words, B is the collection of subsets of R which can be formed from open intervals by any ﬁnite sequence of countable unions, countable intersections, or complements. 3 below). 3. (1) Let A = {A ⊂ I | A is countable, or Ac is countable}. Prove that A is a σ-algebra. (2) Let A = {A ⊂ I | A is a null set, or Ac is a null set}. Prove that A is a σ-algebra. (3) Suppose Aλ is a σ-algebra of subsets of X for each λ in some indexing set Λ. Prove that Aλ A= λ∈Λ is a σ-algebra of subsets of X. (4) Let A be a σ-algebra of subsets of R and suppose I is a closed interval which is in A. 