{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ia2sp10fes

# Solution we have to see f f f g g f and f gg h f h

This preview shows pages 2–4. Sign up to view the full content.

Solution. We have to see f f , f g g f and f g,g h f h , for all functions f,g,h : R R . Since the empty set is a null set, it is clear that f f for all f : R R . Assume f,g : R R and f g . Then the set E = { x : f ( x ) 6 = g ( x ) } is a null set; this is of course the same set as { x : g ( x ) 6 = f ( x ) } , thus g = f a.e. and g f . Alternatively, the deﬁnition of f = g a.e. is symmetric in f,g , thus f = g a.e. if and only if g = f a.e. . In any case f g g f . Finally assume f g,g h . The sets E = { x : f ( x ) 6 = g ( x ) } , F = { x : g ( x ) 6 = h ( x ) } are then null sets. Clearly f ( x ) 6 = h ( x ) implies that either x E or x F , thus { x : f ( x ) 6 = h ( x ) E F , thus a null set. Thus f = h a,e. and f h . 6. Let { f n } n =1 be a sequence of real valued functions on a subset E of R . Deﬁne the following sets: For each triple ( n,m,k ) N 3 deﬁne F n,m,k = { x E : | f n ( x ) - f m ( x ) | < 1 k } . For each pair ( N,k ) N 2 deﬁne G N,k = \ n N,m N F m,n,k . For each k N , deﬁne H k = [ N =1 G N,k . Finally, deﬁne Z = \ k =1 H k . Prove: x Z if and only if the sequence of real numbers { f n ( x ) } converges. Solution. Assume ﬁrst x Z . We prove { f n ( x ) } is a Cauchy sequence of real numbers. For this let ² > 0 be given. Let k N be such that 1 /k < ² . Since x Z , x H k by the deﬁnition of Z . By the deﬁnition of H k , there is N N such that x G N,k . By the deﬁnition of G N,k , x F n,m,k for all n,m N , thus | f n ( x ) - f m ( x ) | < 1 /k < ² if n,m N . It follows the sequence is Cauchy, hence converges. Conversely, let x R be such that { f n ( x ) } converges. Then it is Cauchy and for every k N there is N N such that | f n ( x ) - f m ( x ) | < 1 /k if n,m N , thus there is N N such that x F n,m,k for n,m N , hence x G N,k . This proves that for every k N , there is N N such that x G N,k , thus for every k N , x S N =1 G N,k = H k . Thus x H k for all k , hence x Z . 2

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

View Full Document
7. Let E be a measurable subset of R and let f : E [0 , ] be measurable. Prove that the function 1 /f deﬁned by (1 /f )( x ) = 1 /f ( x ) if 0 < f ( x ) < if f ( x ) = 0 0 if f ( x ) = for x E , is measurable. Solution. I’ll present two solutions. First solution, the obvious one. Let α R . It suﬃces to prove that { x E : 1 /f ( x ) > α } is measurable. We have to distinguish three cases: α > 0, α = 0, α < 0. If α > 0, then 1 /f ( x ) > α if and only if either f ( x ) = 0 (then 1 /f ( x ) = ) or if f ( x ) < 1 . But since 1 /α > 0, the set { f = 0 } ⊂ { f < 1 } . Thus { x E : 1 /f ( x ) > α } = { x E : f ( x ) < 1 } , the latter set being measurable because f is measurable, it follows that { x E : 1 /f ( x ) > α } is measurable if α > 0. Assume next
This is the end of the preview. Sign up to access the rest of the document.
• Spring '11
• Speinklo
• Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Non-measurable set

{[ snackBarMessage ]}

### Page2 / 8

Solution We have to see f f f g g f and f gg h f h for all...

This preview shows document pages 2 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online