By Percus J.K.

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2 as T (x) = (sin 2x)I(0,∞) (x) + (1 + cos x)I(−∞,0] (x), x ∈ R, the B(R), B(R) -measurability of T follows. 2). 1 Measurable transformations 43 Next, measurability of the limit of a sequence of measurable functions is ¯ = R ∪ {+∞, −∞} denote the extended real line and let considered. Let R ¯ ≡ σ B(R) ∪ {+∞} ∪ {−∞} denote the extended Borel σ-algebra on B(R) ¯ R. 5: For each n ∈ N, let fn : Ω → R measurable function. (i) Then, each of the functions supn∈N fn , inf n∈N fn , lim supn→∞ fn , and ¯ -measurable.

Xk ) = xi , (x1 , . . , xk ) ∈ Ω1 = Rk . Then, σ {fi : 1 ≤ i ≤ k} = B(Rk ). To show this, note that any measurable rectangle A1 × . . × Ak can be k written as A1 × . . × Ak = i=1 fi−1 (Ai ) and hence A1 × . . × Ak ∈ σ {fi : 1 ≤ i ≤ k} for all A1 , . . , Ak ∈ R. Since Rk is generated by the collection of all measurable rectangles, B(Rk ) ⊂ σ {fi : 1 ≤ i ≤ k} . Conversely, for any A ∈ B(R) and for any 1 ≤ i ≤ k, fi−1 (A) = R × . . × A × . . × R (with A in the ith position) is in B(Rk ). Therefore, σ {fi : 1 ≤ i ≤ k} = σ {fi−1 (A) : A ∈ R, 1 ≤ i ≤ k} ⊂ B(Rk ).

The subadditivity of µ∗ yields the opposite inequality and so, A1 ∪ A2 ∈ M and hence, M is an algebra. , An ∈ M, An ⊂ An+1 for all n ≥ 1 ⇒ A ≡ n≥1 An ∈ M. 3 The extension theorems and Lebesgue-Stieltjes measures 23 for all n ≥ 2. Then, for all n ≥ 1, Bn ∈ M (since M is an algebra), n ∞ j=1 Bj = An , and j=1 Bj = A. Hence, for any E ⊂ Ω, µ∗ (E) = µ∗ (E ∩ An ) + µ∗ (E ∩ Acn ) = µ∗ (E ∩ An ∩ Bn ) + µ∗ (E ∩ An ∩ Bnc ) + µ∗ (E ∩ Acn ) (since Bn ∈ M) ∗ ∗ ∗ = µ (E ∩ Bn ) + µ (E ∩ An−1 ) + µ (E ∩ Acn ) n = µ∗ (E ∩ Bj ) + µ∗ (E ∩ Acn ) (by iteration) µ∗ (E ∩ Bj ) + µ∗ (E ∩ Ac ) (by monotonicity).