- Suppose X and Y are metric spaces having compact closed balls. Show
is also a metric space which has the closures of balls compact. Here

Let

Show σ

, the smallest σ algebra containing A contains the Borel sets. Hint: Show every open set in a metric space which has closed balls compact can be obtained as a countable union of compact sets. Next show this implies every open set can be obtained as a countable union of open sets of the form U × V where U is open in X and V is open in Y . - Suppose is a measure space which may not be complete. Could you obtain a complete measure space,by simply letting S consist of all sets of the form E where there exists F ∈S such that∪⊆ N for some N ∈S which has measure zero and then let μ= μ
_{1}? Explain. - Let be a σ finite measure space and let f : Ω → [0,∞) be measurable. Define
Verify that A is μ × m measurable. Show that

- For f a nonnegative measurable function, it was shown that
Would it work the same if you used ∫ μ

dt? Explain. - Let be a finite measure space and supposeis a sequence of nonnegative functions which satisfy f
_{n}≤ C independent of n,ω. Suppose also this sequence converges to 0 in measure. That is, for all ε > 0,Show that then

- Explain why for each t > 0,x → e
^{−tx}is a function in L^{1}andThus

Now explain why you can change the order of integration in the above iterated integral. Then compute what you get. Next pass to a limit as R →∞ and show

- Let f= g=
^{−1∕2}if y ∈∪and f= g= 0 if y∪. For which values of x does it make sense to write the integral ∫_{ℝ}fgdy? - Let E
_{i}be a Borel set in ℝ. Show that ∏_{i=1}^{n}E_{i}is a Borel set in ℝ^{n}. - Let be an increasing sequence of numbers inwhich converges to 1. Let g
_{n}be a nonnegative function which equals zero outsidesuch that ∫ g_{n}dx = 1. Now for∈ [0,1) × [0,1) defineExplain why this is actually a finite sum for each such

so there are no convergence questions in the infinite sum. Explain why f is a continuous function on [0,1) × [0,1). You can extend f to equal zero off [0,1) × [0,1) if you like. Show the iterated integrals exist but are not equal. In fact, showDoes this example contradict the Fubini theorem? Explain why or why not.

- Let f : → ℝ be Rieman integrable. Thus f is a bounded function and by Darboux’s theorem, there exists a unique number between all the upper sums and lower sums of f, this number being the Riemann integral. Show that f is Lebesgue measurable and
where the second integral in the above is the Lebesgue integral taken with respect to one dimensional Lebesgue measure and the first is the ordinary Riemann integral.

- Let be a σ finite measure space and let f : Ω → [0,∞) be measurable. Also let ϕ : [0,∞) → ℝ be increasing with ϕ= 0 and ϕ a C
^{1}function. Show thatHint: This can be done using the following steps. Let t

_{i}^{n}= i2^{−n}. Show thatNow this is a countable sum of ℱ×ℬ

measurable functions and so it follows that→X_{[f>t] }is ℱ×ℬmeasurable. Consequently, so is X_{[f>t] }ϕ. Note that it is important in the argument to have f > t. Now observeUse Fubini’s theorem. For your information, this does not require the measure space to be σ finite. You can use a different argument which ties in to the first definition of the Lebesgue integral. The function t → μ

is called the distribution function. - Give a different proof of the above as follows. First suppose f is a simple
function,
where the a

_{k}are strictly increasing, ϕ= a_{0}≡ 0. Then explain carefully the steps to the following argument. - Give another argument for the above result as follows.
and now change the variable in the last integral, letting ϕ

= t. Justify the easy manipulations.

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