Question

real analysis question.

Answer 1

@eliassaab

Answer 2

Where is it?

Answer 3

the question is prove proposition 7 of page 9 in http://www.math.ucla.edu/~tao/resource/general/131ah.1.03w/week9.pdf

Answer 4

could you verify my proof?

Answer 5

ok so let P*P' be the common refinement of P and P'

Answer 6

by prior facts, f would also be piecewise continuous on P*P' and that every J (I denote generalized intervals by capital letters) element of P*P' is a subset of a gen. interval in P

Answer 7

by definition,
\[p.c. \int\limits_{[P*P']}f=\sum_{X \in P*P'}c_x|X|\]
where c_x is the constant value of f on the gen. interval X and |X| is the length of X.

Answer 8

now consider the set
\[P_{J}=\left\{ X \in P*P':J \in P, X \subseteq J \right\}\]
I know that for every gen. int. J,J' of P, P_J and P_J' are disjoint. suppose on the contrary it is not, then there exists an element j (this time its not a gen. int., I use small letters for single elements) in P_J which is also in P_J' so that x is in J and J' which contradicts the definition of a partition.

Answer 9

now by summation laws on finite sets, the RHS is then:
\[\sum_{J \in P}\sum_{X \in P_J}c_x|X|\]

Answer 10

now we know that c_X=c_J whenever X is a subset of J. suppose on the contrary, c_X is not equal to c_J even when X is in the generalized interval J, then f would have two values at J, but then f is piecewise constant with respect to P, and J is a gen. interval on P, so f must be constant on J, a contradiction.

Answer 11

now the above is equal to
\[\sum_{J \in P}\sum_{X \in P_J}c_J|X|\]
since X is a subset of J here

Answer 12

Now notice that P_J will be a partition of J. to prove this, consider any element j in J. Since J is in a partition P of I, then x must be an element of I by def, and that x will only be contained in J or in some of its subsets. Now suppose on contrary that P_J is the set of all X in partition P which is subset of J, and j element of J doesn't appear exactly once in P_J. Now since P_J is just a disjoint union of X which is in partition P*P' of I, and that j is an element of I which is in J only in the partition P, then it must appear exactly once in the partition P*P' only. In particular, it must appear once in one of the gen. intervals in partition P*P' and is subset of J. But that is just P_J, hence we have a contradiction.

So P_J is a partition of J

Answer 13

now by a prior theorem,
\[|J|=\sum_{X \in P_J}|X|\]
\[c_J|J|=c_J \sum_{X \in P_J}|X|=\sum_{X \in P_J}c_J|X|\]
but this is true for any J element of P, hence

Answer 14

\[\sum_{J \in P} \sum_{X \in P_J}c_J|X|=\sum_{J \in P}c_J|P|=p.c. \int\limits_{P}f\]
hence
\[p.c. \int\limits_{P*P'}f=p.c. \int\limits_{P}f\]
now I used a similiar argument to show that:
\[p.c.\int\limits_{P*P'}f=p.c.\int\limits_{P'}f\]
the result follows.

Answer 15

@eliassaab , sorry I haven't typed the whole solution last night, my computer was lagging....