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4. Let \(A\) be a dense subset of \(\mathbb{R}\). Show that \(f\) is measurable if \(\{x:\ f(x)>a\}\) is a measurable set for all \(a\in A\).
5. Show that 1. The sum and product of two simple functions are simple. 2. \(\chi_{A\cap B}=\chi_{A}\cdot\chi_{B}\). 3. \(\chi_{A\cup B}=\chi_{A}+\chi_{B}-\chi_{A\cap B}\). 4. \(\chi_{A^{c}}=1-\chi_{A}\).
6. (Tchebychev Inequality). Suppose \(f\geq 0\) and integrable. If \(\lambda>0\) and \(E_{\lambda}=\{x:\ f(x)>\lambda\}\), then show that \[m(E_{\lambda})\leq\frac{1}{\lambda}\int f\,dx.\]
7.
1. Give an example of a function \(f:\mathbb{R}\to\mathbb{R}\) which is integrable, but \(f^{2}\) is not.
2. Show that a measurable function \(f\) is integrable if and only if \(|f|\) is integrable. Give an example of a nonintegrable function whose absolute value is integrable.
8. Suppose \(f\) is Riemann integrable on the closed interval \([a,b]\). Then show that \(f\) is measurable and \[(R)\int_{[a,b]}f\,dx=(L)\int_{[a,b]}f\,dx,\] where the integral on the left-hand side is the Riemann Integral, and the one on the right-hand side is the Lebesgue integral.
9. Let \(\mu\) be the counting measure on \(\mathbb{N}\). Interpret Fatou’s lemma and the monotone and dominated convergence theorems as statements about infinite series.
10. If \(f_{n},g_{n},f,g\) are all integrable functions with \(f_{n}\to f\) and \(g_{n}\to g\) a.e., \(|f_{n}|\leq g_{n}\), and \(\int g_{n}\,dx\to\int g\,dx\), then show that \(\int f_{n}\,dx\to\int f\,dx\).
11. Suppose \(f_{n},f\in L^{1}\) and \(f_{n}\to f\) a.e.. Then show that \(\int|f_{n}-f|\,dx\to 0\) if and only if \(\int|f_{n}|\,dx\to\int|f|\,dx\).
12. Suppose \(f\geq 0\); let
\[\mu(E)=\int_{E}f\,dm\]
for a measurable set \(E\).
1. Show that \(\mu\) is a measure.
2. Show that for any \(g\geq 0\) \[\int g\,d\mu=\int fg\,dm.\] Hint: First suppose \(g\) is simple.
13. Show that \(f(x)=\frac{\ln x}{x^{2}}\) is Lebesgue integrable over \([1,\infty)\) and that \(\int fdx\)=1.
14. Establish the Riemann-Lebesgue Theorem: if \(f\) is an integrable function on \((-\infty,\infty)\) then
\[\lim_{n\to\infty}\int_{-\infty}^{\infty}f(x)\cos(nx)\,dx=0.\]
Hint: The theorem is easy if \(f\) is a simple function, then use Theorem 2.4 in [50], p. 71.
15. Let \(f\) be integrable over \((-\infty,\infty)\), then show that
1. \[\int f(x)\,dx=\int f(x+t)\,dx.\]
2. Let \(g\) be a bounded measurable function. Then show that \[\lim_{t\to 0}\int_{-\infty}^{\infty}|g(x)[f(x)-f(x+t)]|\,dx=0.\]
16.
1. Let \(\{f_{n}\}\) be a sequence of real-valued measurable functions. If \(\{f_{n}\}\) converges to \(f\) in measure, show that \(\{f_{n}\}\) is a Cauchy sequence in measure.
* Show that if a sequence \(\{f_{n}\}\) of integrable functions converge to \(f\) in \(L^{1}\), then \(\{f_{n}\}\) converges \(f\) in measure. Is the converse true? Note: We say \(\{f_{n}\}\) is _Cauchy in measure_ if for every \(\epsilon>0\), \[m(\{x:\ |f_{n}(x)-f_{m}(x)|>\epsilon\})\to 0\ \ \ \mbox{as}\ m,n\to\infty\] and we say \(\{f_{n}\}\)_converges in measure_ to a measurable function \(f\) if for every \(\epsilon>0\) \[m(\{x:\ |f_{n}(x)-f(x)|>\epsilon\})\to 0\ \ \ \mbox{as}\ n\to\infty.\]
17. Compute the following limits and justify the calculations:
* \(\lim_{n\to\infty}\int_{0}^{\infty}[1+(x/n)]^{-n}\sin(x/n)dx\).
* \(\lim_{n\to\infty}\int_{0}^{\infty}n\sin(x/n)[x(1+x^{2})]^{-1}dx\).
* \(\lim_{n\to\infty}\int_{a}^{\infty}n(1+n^{2}x^{2})^{-1}dx\).
18. Show that \(\int_{0}^{\infty}x^{2n}e^{-x^{2}}dx=\frac{(2n)!}{2^{2n}n!}\cdot \frac{\sqrt{\pi}}{2}\) holds true for \(n=0,1,2,\ldots\)
Hint: Use induction on \(n\) and the fact that \(\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}\) (Euler’s formula).
19. Show that for \(a>0\), \(\ \int_{-\infty}^{\infty}e^{-x^{2}}\cos(ax)\,dx=\sqrt{\pi}e^{-a^{2}/4}\).
Hint: Use the problem above.
20. Let \(M\) be an arbitrary metric space and \(f:\mathbb{R}\times M\to\mathbb{R}\) a function. Assume that \(f(-,t)\) is a measurable function for each \(t\in M\) and \(f(x,-)\) is a continuous function for each \(x\in\mathbb{R}\). Assume also that there exists an integrable function \(g\) such that for each \(t\in M\) we have \(|f(x,t)|\leq g(x)\) for almost all \(x\in\mathbb{R}\). Show that the function \(F:M\to\mathbb{R}\) defined by
\[F(t)=\int_{\mathbb{R}}f(x,t)\ \ dx\]
is a continuous function.
### 2.8 Banach-Tarski Paradox
Consider the closed unit ball in three dimensions,
\[B=\{(x,y,z)\in\mathbb{R}^{3}:\quad x^{2}+y^{2}+z^{2}\leq 1\}.\]
Can \(B\) be decomposed into a finite number of parts and reassembled into two balls identical to the original \(B\)? Our first response will be “No!” (Figure 2.28).
The intuition one has about volumes of everyday objects in the physical world makes us think this way. Doubling the ball by dividing it into parts and moving the parts around by rigid motions (i.e., rotations and translations) without any stretching, bending, or adding new points seems to be impossible. However, perhaps the intuition we rely on is not applicable to some subsets in the world of \(\mathbb{R}^{3}\). In fact, in a paper published in 1924, Stefan Banach and Alfred Tarski gave a construction of such a paradoxical decomposition. The ordinary “volume” of a subset of \(\mathbb{R}^{3}\) or \(\mathbb{R}^{n}\) is its three-dimensional or \(n\)-dimensional Lebesgue measure, respectively. It is possible that not all subsets of \(\mathbb{R}^{n}\) are measurable; in fact the pieces in the Banach-Tarski paradox are not measurable. Therefore, we can be assured that the rules concerning volume are actually not violated here. We will not prove the Banach-Tarski decomposition in this section. A proof of it and its consequences are given in a book by Wagon [53] in detail. In the following we answer the question of whether or not all subsets of \(\mathbb{R}^{n}\) are measurable. For this we need to recall the axiom of choice and the definition of measure.
### Axiom of Choice
The Axiom of Choice appears in many proofs in mathematics. For a long time mathematicians discussed among themselves about the use of Axiom of Choice and seemingly contradictory results it produced. Now most seem to agree that
Figure 2.28: Banach–Tarski paradoxthe advantages of the Axiom of Choice outweigh the disadvantages.
Suppose \(I\) is an index set and \(\{A_{i}\}_{i\in I}\)is a collection of nonempty sets. We call \(\psi\) a _choice function_ if \(\psi:I\to\bigcup_{i\in I}A_{i}\) is such that \(\psi(i)\in A_{i}\).
**The Axiom of Choice:** For every collection of nonempty sets there exists a choice function.
Note that the Axiom of Choice (AC) is a nonconstructive assertion of existence. It postulates the existence of certain objects without giving any indication of how to find these objects. There are various logically equivalent statements such as Hausdorff Mimality Principle, Zorn’s Lemma, and Well-Ordering Principle, see the end of Section (1.1) for more information.
### Construction of a Non-Measurable Set
Are all subsets of \(\mathbb{R}^{n}\) measurable? In this section we answer this question for \(n=1\) by constructing a subset of \(\mathbb{R}\) which is not measurable. The construction of a non-measurable set uses the Axiom of Choice. First we recall the concept of measure on \(\mathbb{R}\).
**Definition 102**.: We call a non-negative function \(m\) a _measure_ on the family of \(E\subset\mathbb{R}\) if
* \(m(E)=b-a\) if \(E=[a,b],\ a\leq b\).
* \(m(E)=\sum_{n=1}^{\infty}m(E_{n})\), whenever \(E=\bigcup_{n=1}^{\infty}E_{n}\) and the sets \(E_{n}\) are disjoint.
* \(m(E+h)=m(E)\) for every \(h\in\mathbb{R}\).
The construction of a non-measurable set \(\mathcal{M}\) rests on a simple equivalence relation among real numbers in \([0,1]\). We say \(x\) is equivalent to \(y\), and write this as
\[x\sim y\quad\text{ whenever }\quad x-y\quad\text{ is rational.}\]
Note that this is an equivalence relation since the following properties hold:
* \(x\sim x\) for every \(x\in[0,1]\).
* if \(x\sim y\), then \(y\sim x\).
* if \(x\sim y\) and \(y\sim z\), then \(x\sim z\).
Two equivalent classes are either disjoint or coincide and furthermore \([0,1]\) is the disjoint union of these equivalence classes \(E_{\alpha}\). In other words,
\[[0,1]=\bigcup_{\alpha}E_{\alpha}.\]
Now we construct a set \(\mathcal{M}\) by choosing exactly one element \(x_{\alpha}\) from each \(E_{\alpha}\), and setting \(\mathcal{M}=\{x_{\alpha}\}\). Note that the construction of the set \(\mathcal{M}\) is possible because of the Axiom of Choice.
**Theorem 119** (G. Vitali: Non-measurable set).: _The set \(\mathcal{M}\) is not measurable._
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