(New page: 9b) Let <math>[a,b] \subset [0,1]</math>. Since F is continuous, <math>F([a,b])</math> is compact, thus <math>\exists \alpha , \beta \in [a,b]</math> such that <math>F(\alpha) \leq F(x) ...) |
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| + | 9a) | ||
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| + | Let E be measurable. Then, <math>\exists</math> an <math> F_{\sigma}</math> set B and a set of measure zero Z such that | ||
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| + | <math> E = B\cup Z</math> (disjoint union). | ||
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| + | Then <math>F(E)=F(B)\cup F(Z)</math>. | ||
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| + | Now F is A.C., thus, F(Z) has measure zero. To see this, realize if <math>\epsilon > 0</math> then let <math>\delta</math> be number corresponding to <math>\epsilon</math> in the definition of A.C. Take a collection of open intervals covering Z the sum of whose lenghts is <math><\delta</math>, then the sum of the lengths of the image of these intervals under F will be <math>< \epsilon</math>. | ||
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| + | Also, since B is an <math>F_{\sigma}</math> it is a countable union of closed sets. Each closed set can be written as a countable union of compact sets, so B can be written as the countable union of compact sets. That is,<math> B = \cup_n C_n ; \ C_n</math> are compact. | ||
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| + | Then <math>F(B) = F( \cup_n C_n) = \cup_n F(C_n)</math> which is the countable union of compact hence closed sets, so F(B) is an <math>F_{\sigma}</math>. | ||
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| + | So F(E) is the union of a two measurable sets, hence measurable. | ||
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9b) | 9b) | ||
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= \sum_n m(F([a_n,b_n])) \leq \sum_n \int_{[a_n,b_n]} |f(t)| dt | = \sum_n m(F([a_n,b_n])) \leq \sum_n \int_{[a_n,b_n]} |f(t)| dt | ||
| − | = \int_G |f(t)| dt </math> | + | = \int_G |f(t)| dt </math> since the intervals are nonoverlapping. |
Finally for any E, let <math>G_k</math> be a collection of open sets <math>\searrow E</math>. Then <math>F(G_k) \searrow F(E)\Rightarrow m(F(G_k)) \searrow m(F(E))</math> so, | Finally for any E, let <math>G_k</math> be a collection of open sets <math>\searrow E</math>. Then <math>F(G_k) \searrow F(E)\Rightarrow m(F(G_k)) \searrow m(F(E))</math> so, | ||
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= \int_0^1 lim_{k \rightarrow \infty} |f(t)|\chi_{G_k} dt = \int_0^1 |f(t)|\chi_{E} dt = \int_E |f(t)| dt \Rightarrow </math> claim. | = \int_0^1 lim_{k \rightarrow \infty} |f(t)|\chi_{G_k} dt = \int_0^1 |f(t)|\chi_{E} dt = \int_E |f(t)| dt \Rightarrow </math> claim. | ||
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| + | Dat & Ben W | ||
Latest revision as of 14:20, 22 July 2008
9a)
Let E be measurable. Then, $ \exists $ an $ F_{\sigma} $ set B and a set of measure zero Z such that
$ E = B\cup Z $ (disjoint union).
Then $ F(E)=F(B)\cup F(Z) $.
Now F is A.C., thus, F(Z) has measure zero. To see this, realize if $ \epsilon > 0 $ then let $ \delta $ be number corresponding to $ \epsilon $ in the definition of A.C. Take a collection of open intervals covering Z the sum of whose lenghts is $ <\delta $, then the sum of the lengths of the image of these intervals under F will be $ < \epsilon $.
Also, since B is an $ F_{\sigma} $ it is a countable union of closed sets. Each closed set can be written as a countable union of compact sets, so B can be written as the countable union of compact sets. That is,$ B = \cup_n C_n ; \ C_n $ are compact.
Then $ F(B) = F( \cup_n C_n) = \cup_n F(C_n) $ which is the countable union of compact hence closed sets, so F(B) is an $ F_{\sigma} $.
So F(E) is the union of a two measurable sets, hence measurable.
9b)
Let $ [a,b] \subset [0,1] $. Since F is continuous, $ F([a,b]) $ is compact, thus $ \exists \alpha , \beta \in [a,b] $ such that $ F(\alpha) \leq F(x) \leq F(\beta) \ \forall \ x \in [a,b] $ or $ F(\alpha) \geq F(x) \geq F(\beta) \ \forall \ x \in [a,b] $ .
Then $ m(F(E))\leq |F(\alpha)-F(\beta)| = |\int_{\alpha}^{\beta} f(t) dt| \leq \int_{\alpha}^{\beta} |f(t)| dt = \int_{[a,b]} |f(t)| dt $
So the claim holds for a compact interval $ [a,b]. $
Now let G be any open set. Then we can write G as a countable union of nonoverlapping compact intervals:
$ G = \cup_{n=1}^\infty [a_n,b_n] $
Then $ m(F(G)) = m(F(\cup_{n=1}^\infty [a_n,b_n]))=m(\cup_{n=1}^\infty F([a_n,b_n])) $
$ = \sum_n m(F([a_n,b_n])) \leq \sum_n \int_{[a_n,b_n]} |f(t)| dt = \int_G |f(t)| dt $ since the intervals are nonoverlapping.
Finally for any E, let $ G_k $ be a collection of open sets $ \searrow E $. Then $ F(G_k) \searrow F(E)\Rightarrow m(F(G_k)) \searrow m(F(E)) $ so,
$ m(F(E)) = lim_{k \rightarrow \infty} m(F(G_k)) \leq lim_{k \rightarrow \infty} \int_0^1 |f(t)|\chi_{G_k} dt = \int_0^1 lim_{k \rightarrow \infty} |f(t)|\chi_{G_k} dt = \int_0^1 |f(t)|\chi_{E} dt = \int_E |f(t)| dt \Rightarrow $ claim.
Dat & Ben W
