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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 46 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 8. Tuesday, Mar 3: Cauchys inequality and Lebesgue measure I first discussed the definition of preHilbert and Hilbert spaces and proved Cauchys inequality and the parallelogram law. This can be found in all the lecture notes and many other places so I will not repeat it here. Another nice source is the book of G.F. Simmons, Introduction to topology and modern analysis. I like it but I think it is out of print. In case anyone is interested in how to define Lebesgue measure from where we are now and I may have time to do this later we can just use the integral as I outlined on Tuesday. First, we define locally integrable functions. Thus f : R C is locally integrable if (8.1) F [ N,N ] = f ( x ) x [ N,N ] 1 ( R ) N. x if  x  > N L For example any continuous function on R is locally integrable. Lemma 4. The locally integrable functions form a linear space. Proof. Follows from the linearity of L 1 ( R ) . Definition 5 . A set A R is measurable if its characteristic function A is locally integrable. A measurable set A has finite measure if A L 1 ( R ) and then (8.2) ( A ) = N is the Lebesgue measure of A. If A is measurable but not of finite measure then ( A ) = by definition. We know immediately that any interval ( a,b ) is measurable (whether open, semi open or closed) and has finite measure if and only if it is bounded then the measure is b a. Some things to check: Proposition 13. The complement of a measurable set is measureable and any countable union of measurable sets is measurable. Proof. The first part follows from the fact that the constant function 1 is locally integrable and hence R \ A = 1 A is locally integrable if and only if A is locally integrable. Notice the relationship between characteristic functions and the sets they define: (8.3) A B = max( A , B ) , A B = min( A , B ) . S If we have a sequence of sets A n then B n = A k is clearly an increasing sequence of sets and k n (8.4) B n B , B = A n n is an increasing sequence which converges pointwise (at each point it jumps to 1 somewhere and then stays or else stays at . ) Now, if we multiply by [ N,N ] then (8.5) f n = [ N,N ] B n B [ N,N ] is an increasing sequence of integrable functions assuming that is that the A k s are measurable with integral bounded above, by 2 N. Thus by our monotonicity S theorem the limit is integrable so B is locally integrable and hence n A n is measurable. [ [ 47 LECTURE NOTES FOR 18.102, SPRING 2009 Proposition 14. The (Lebesgue) measurable subsets of R form a collection, M , of the power set of R , including and R which is closed...
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This note was uploaded on 11/08/2011 for the course PHY 18.102 taught by Professor Staff during the Spring '09 term at MIT.
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