ia2sp10h9s

# The complement of a countable set being dense we can

• Notes
• 7

This preview shows page 3 - 6 out of 7 pages.

increasing function is countable. The complement of a countable set being dense, we can find sequences { c n } , { d n } with the following properties: (a) a < c n +1 < c n , c 1 < d 1 , d n < d n +1 < b for all n N (b) lim n →∞ c n = a , lim n →∞ d n = b . (The sequence { c n } decreases to a , { d n } increases to b . (c) f is continuous at c n , d n for all n N . 3

Subscribe to view the full document.

Let J n = ( c n +1 , c n ), K n = ( d n , d n +1 ) for n N ; H = ( c 1 , d 1 ). Let also ² > 0 be given. Because f is continuous at c n , there is δ n > 0 such that | x - c n | < δ n implies | f ( x ) - f ( c n ) | < ²/ 2 n +2 . In particular, f ( c n + δ n / 2) - f ( c n - δ n / 2) = f ( c n + δ n / 2) - f ( c n ) + f ( c n ) - f ( c n - δ n / 2) < ² 2 n +1 . Similarly, there is η n > 0 such that | x - d n | < η n implies | f ( x ) - f ( d n ) | < ²/ 2 n +2 . In particular, f ( d n + η n / 2) - f ( d n - η n / 2) = f ( d n + η n / 2) - f ( d n ) + f ( d n ) - f ( d n - η n / 2) < ² 2 n +1 . Let L n ( c n - δ n / 2 , c n + δ n / 2), M n ( d n - η n / 2 , d n + η n / 2) for n N . Since this may look a bit complicated, I will insist that the idea is quite simple and natural; draw a picture! The family of open intervals { J n } ∪ { K n } ∪ { L n } ∪ { M n } ∪ { H } is a countable family covering ( a, b ) and if we add all the ˜ values of these intervals we notice X n =1 ˜ ( J n ) = lim n →∞ n X k =1 ( f ( c k ) - f ( c k +1 )) = f ( c 1 ) - lim n →∞ f ( c n +1 ) = f ( c 1 ) - f ( a +) . X n =1 ˜ ( K n ) = lim n →∞ n X k =1 ( f ( d k +1 ) - f ( d k )) = lim n →∞ f ( d n +1 ) - f ( d 1 ) = f ( b - ) - f ( d 1 ) . ˜ ( H ) = f ( d 1 ) - f ( c 1 ) . X n =1 ˜ ( L n ) = X n =1 f c n + δ n 2 - f c n - δ n 2 ¶¶ < X n =1 e 2 n +1 = ² 2 . X n =1 ˜ ( M n ) = X n =1 f d n + η n 2 · - f d n - η n 2 ·· < X n =1 e 2 n +1 = ² 2 . Because we have a countable cover of I , the sum of the expressions on the left of the first of all these equali- ties/inequalitites dominates μ * ( I ); adding up we thus get μ * ( I ) < f ( b - ) - f ( a +) + ². Since ² > 0 is arbitrary, μ * ( I ) f ( b - ) - f ( a +) follows, establishing Claim 2. Claim 3. If I = ( a, b ], -∞ ≤ a < b < , then (7) μ * ( I ) = f ( b +) - f ( a +) Assume first that a > . We can proceed exactly as we did in proving the first inequality in (4), with two difference. The countable cover { J n } is defined in the same way in terms of ² , but then we use the fact that [ a + η, b ] is compact to extract a finite subcover of [ a + η, b ]. That is the first difference; that b - η does not appear; it is replaced by b . The second difference is that where we used f ( b - η ) f ( d n k ), now we use that f ( b +) f ( d n k ). Everything else is the same, and we end with μ * ( I ) f ( b +) - f ( a +). For the converse of this inequality, select any c ( a, b ) at which f is continuous. For a small δ consider the open intervals ( a, c ) , ( c - δ, c + d ) , ( c, b + δ ). These intervals cover I ; because μ * is an outer measure and by Claim 2: μ * ( I ) μ * (( a, c ))+ μ * (( c - δ, c + δ ))+ μ * (( c, b + δ )) = f ( c - ) - f ( a +)+ f (( c + δ ) - ) - f (( c - δ )+)+ f (( b + d ) - ) - f ( c +) . Using the continuity of f at c , letting δ 0, we get μ * ( I ) f ( b +) - f ( a +), establishing Claim 3 . With this we can return to where we were on hold, namely at (3). With the notation as defined there, we have (8) μ * ( I 1 n ) + μ * ( I 2 n ) = μ * ( I n ) = f ( b n - ) - f ( a n +) 4
for all n N . (The last equality is, of course, just claim 2). Let I n = ( a n , b n ). If one of I 1 n , I 2 n is empty, then the other one is I n and (8) is trivially true. Otherwise, a n < a < b n and I 1 n = ( a, b n ), I 2 n = ( a n , a ] and by claims 2, 3

Subscribe to view the full document.

• Spring '11
• Speinklo
• CN, Lebesgue measure, open intervals

{[ snackBarMessage ]}

## Get FREE access by uploading your study materials

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
###### "Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern