V.I. Levenshtein. Binary codes capable of correcting deletions, insertions
and reversals. Soviet Physics Doklady, 6:707-710, 1966.
 B. Lewin. Genes VII. Oxford University Press, 1999.
 M. Li, B. Ma, and L. Wang. Finding si
318 Measure and Outer Measure [Chap 12
Inner measure was important historically because the measur-
ability of a set was originally characterized using both inner and
outemeasure. hi the historicai WWW
dened for bounded subsets of Wor such sets the deni
Sec. 1] Measure Spaces
2. Proposition: IfEi e (B, ,uEl < co and E D EH1, then
PmoSet E= LThen
E1 = E U U (Ei~Ei+1)a
and this is a disjoint union. Hence
ETET+1 U (ET~E
322 Measure and Outer Measm [Chap. 12
result for outer measure is true even if the sets A are only assumed
to be measurable (see Problems 2 and 46e).
37. Lemma: Let (A) be a disjoint sequence ofsets in 3. Then
(E n 91A,.) = :21 ME n A).
SELL 2] The Extension Theorem 297
and this is a countable disjoint union of sets in Q, and so we have
173 = 2 Mi n B)
This extension procedure not only extends u to a measure on the
Sec. 3] Integration 265
If [a (a < 00, then the set A = cfw_x a E: go(x) > 0 is a measurable
set of nite measure. Let M be the maximum of go, let 6 be a given
=-_ cfw_x g Esz(x) > (1 )(p(x) for all k _>_ n.
Then (AH) is an inc
Sec. 4] Product Measures 311
22. Let h and g be integrable functions on X and Y, and dene
f(x, y) = h(x)g( y). Then s integrable on X x Y and
XxY X Y
(Note: We do not need to assume that p and v are anite.)
ow a e . e . . .
SBLA_]_ELQdJJ1 Measures 307
Proof: By Proposition 6 there is a set F in (Rm, such that E C F
and ,u x v(F) = 0. It follows from Lemma 16 that for almost all x we
is complete. I
#4:;meng Qperajors 313
starting with the semialgebra of rectangles of the form R = A1 x x A,l
and MR): n y; A;, and using the Carathodory extension procedure.
Show that if we identify tXlx-xXp)x(Xp+1x-~XXHJ with
(X1 X X XuLlhenfi X X Fp) X mp+1 X X u)
Then h1, h2, gl, and gz are in the unit balls of [5101), 201), [5(v), and
Li(v), respectively. Also,
ll hkg = lhlkgwhz kgzyi
by the Holder inequaty for u = 1/1 and u* = 1/(1 Hence
If hkg S mliB1)W2'2)lA
and the result follows by
14. Lemma: Let cfw_(Ai x Bi) be a countable disjoint collection of
measurable rectangles whose union is a measurable rectangle A x B.
1(A x B) = 2 MA,- x 3,).
belongs to exactly one
Proof: By the denition of u* there is a sequence <A,-> from (i
such that E C U A,- and
Set A - UA,.Then pJA $211M, _ ZpA,.
To prove the second statement, we note that for each positive
Sec. 4] Product Measures 309
of f can be determined by iterated integration using the following
20. Theorem (Tonelli): Let (,X (i, j) and (Y, (B, v) be two a -nite
Xx Y. Then
i. For almost all x the function fx dened by fx( y) = f
300 Meaane and Outer Measure IChap. 12
Since (a, b] is the intersection of the sets (a, b + l/n], Proposition
11.2 implies that
,u(a, b] = lim u(a, b + i],
m): an<b+%) = F(b+).
Thus a cumulative distribution function
Sea 1] Baire Sets and gore! Sets 333
3. Lemma: Every bounded set is contained in a compact Ga. Every
o-bounded set E is contained in a a-compact open set 0. If E is
4I:emmaJLet61bearingofsets, andletl cfw_Ez E861.
Sec.*GJ lunerMeasuLe 11]
26. Proposition: Let g e [5 and k a E, with l/q + 1/r > 1. Then
f (x) = L W - y)g(y) dy
_i;denedjaLalmast all 2c and
llfllp S llkllr llgllq,
1 1 1
where = - + L
P CI r
W by Pr
Sec. 2] The Extension Theorem 293
Since 6 was an arbitrary positive number,
I . I. _I
51 15' I. 1 *
Proof: Let E be an arbitrary set of nite outer measure and e a
Z ,uAi < )u*E + 6.
By the additivity of
_Sec._*Z]_ExtensiQn by Sets of Measure Zero 325
u*X = 2, *0 = O, and ,u*E = l ifE is not X or Q).
a. Calculate 11*.
b. What are the measurable subsets of X?
c. Show that there is a nonmeasurable set E with ME = ,u*E.
d. Show that the rst and third
Sec, 9| Hausdorff Measures 329
An outer measure ,u* on subsets of a metric space X with the
property that ,u*(A U B) = MA + ,u*B whenever p(A, B) > 0 is
ii I 3 l E F 3? .
51. Prove Proposition 41. [Let F be the set of f
Sec. 2] Measurable Functions 259
b.- If u is a measure on (R, dene [:l on (B by [1E = uE if E 8 (R and
[E = 00 ifE 3 OT. Then [1 is a measure on (B.
if E e (R. Then g is also a measure on (B
298 MeasuLLandMQLMeasuLe [Chan 12
extension to a measure on the algebra (1 generated by C if the follow-
ing conditions are satised:
sets in C, then uC = Z pCi.
ii. Ifa set C in (3 is the union ofa
CHAPTER 12. PROBLEMS
The well-known variant of sorting by transpositions is sorting by transpositions
p(i,i + l,i + 2) where the operation is an exchange of adjacent elements. A
simple bubble-sort algorithm solves this problem for linear permutations.
CHAPTER 11. COMPUTATIONAL PROTEOMICS
Figure 11.8: Aligning spectra. The shared peaks count reveals only D(0) = 3 matching peaks
on the main diagonal, while spectral al
CHAPTER 12. PROBLEMS
The simplest heuristic for the Shortest Superstring Problem is the GREEDY algorithm: repeatedly merge a pair of strings with maximum overlap until only one
string remains. Tarhio and Ukkonen, 1988  defined the
 S. Hannenhalli, C. Chappey, E. Koonin, and RA. Pevzner. Genome sequence
comparison and scenarios for gene rearrangements: A test case. Genomics,
 S. Hannenhalli and RA. Pevzner. Transforming men into mice (pol
 M. Muzio, A.M. Chinnaiyan, EC. Kischkel, K. O'Rourke, A. Shevchenko,
J. Ni, C. Scaffidi, J.D. Bretz, M. Zhang, R. Gentz, M. Mann, RH. Krammer,
M.E. Peter, and V.M. Dixit. FLICE, a novel FADD-homologous ICE/CED3-like protease, is rec
CHAPTER 10. GENOME REARRANGEMENTS
Genomic_Sort (II, F)
Construct the graph G = (2(11, F)
Close all IlT-paths in simple components of (7(11, F) (lemma 10.32)
Close all but one IIF-path in components having more than one ITF-path inside them
while G con
CHAPTER 10. GENOME RE ARRANGEMENTS
Lemma 10.31 For every two unoriented UT-paths located on the same chromosome, there exists an oriented gray edge that joins these paths into a TYT-path.
In a search for an optimal capping, we first ignore the term /(
since B is measurable. Thus
and the lemma follows by letting 6 + 0. I
. .1 l ['
32. Proposition: Let
41. Let ,u be a measure on an algebra (i, and let E be a u* measurable
set. Show that the measure it dened in Theorem 38 has the property that
Industrial hygiene also referred as occupational hygiene are techniques for the
anticipation, evaluation, control and prevention, and recognition of environmental hazards that
may possibly cause illness, injur
Ikea was founded by Ingvar Kamprad and it was first established in 1943 when he was
only 17 years old. He used a cash gift from his father to build a company known as Ikea. At first,
Ikea was a catalog that se
The United States has become very concerned with their cyber security. In the previous
years article entitled Congressional Action on Cybersecurity Would Send Strong Message to
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