Midterm1 - Analysis Midterm 1 Solutions 1 Let(Ω F μ be a...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Analysis Midterm 1 Solutions 1. Let (Ω , F , μ ) be a measure space on which f, f 1 , f 2 , . . . are nonnegative integrable functions. Show that if f n → f almost everywhere and R f n dμ → R fdμ then R | f n- f | dμ → 0. Solution Notice that 0 6 ( f n- f )- 6 f since f, f n are nonnegative. The Dominated Convergence Theorem therefore implies that R ( f n- f )- dμ → since f n → f μ-ae and the dominating function f is integrable. Now observe that Z | f n- f | dμ = Z ( f n- f ) dμ + 2 Z ( f n- f )- dμ tends to 0 since on the right the first summand does so (given) and the second summand does so (shown). [This result is often called Sheff´ e’s lemma .] 2. Let (Ω , F , μ ) be a probability space. For Z ⊂ Ω define μ * ( Z ) = sup { μ ( A ) : F 3 A ⊂ Z } μ * ( Z ) = inf { μ ( B ) : Z ⊂ B ∈ F} define G = { Z ⊂ Ω : μ * ( Z ) = μ * ( Z ) } and for Z ∈ G write ν ( Z ) = μ * ( Z ) = μ * ( Z ). Prove that: (i) Z ∈ G if and only if for each ε > 0 there exist A, B ∈ F such that A ⊂ Z ⊂ B and μ ( B- A ) < ε ; (ii) G is a σ-algebra (containing F ) on which ν is a measure (extending μ ). Solution (i) Let ε > 0. ( ⇒ ) If Z ∈ G then there exist F 3 A ⊂ Z such that ν ( Z )- ε/ 2 < μ ( A ) and Z ⊂ B ∈ F such that μ ( B ) < ν ( Z ) + ε/ 2; now μ ( B- A ) = μ ( B )- μ ( A ) < ε . ( ⇐ ) Let A, B ∈ F satisfy A ⊂ Z ⊂ B and μ ( B- A ) < ε : then μ * ( Z ) > μ ( A ) > μ ( B )- ε > μ * ( Z )- ε so that μ * ( Z...
View Full Document

{[ snackBarMessage ]}

Page1 / 4

Midterm1 - Analysis Midterm 1 Solutions 1 Let(Ω F μ be a...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online