PMATH
PMATH450_S2015.pdf

# Then u f p n x k 1 sup f x k 1 x k x k x k 1 and l f

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Then U ( f, P ) = n X k =1 sup f [ x k - 1 ,x k ] ( x k - x k - 1 ) and L ( f, P ) = n X k =1 inf f [ x k - 1 ,x k ] ( x k - x k - 1 ) are the upper and lower Riemann sums of f . If Q is a refinement of P , then U ( f, P ) U ( f, Q ) and L ( f, P ) L ( f, Q ). If P 1 and P 2 are two partitions, take Q to be their common refinement, so U ( f, P 1 ) U ( f, Q ) L ( f, Q ) L ( f, P 2 ). If inf P U ( f, P ) = sup P L ( f, P ), we say f is Riemann integrable, and we write R R b a f = inf P U ( f, P ). Theorem. Every continuous function is Riemann integrable. Example. Take f ( x ) = χ Q ( x ) = ( 1 x 2 Q 0 x / 2 Q . This is discontinuous everywhere, and not integrable on [ a, b ], since U ( f, P ) = b - a and L ( f, P ) = 0. Example. Consider Q \ [0 , 1] = { r n } 1 n =1 . Define f n ( x ) = ( 1 x = r 1 , . . . , r n 0 otherwise . f n is Riemann integrable, with R R 1 0 f n = 0 for all n . But notice that f n ( x ) ! χ Q \ [0 , 1] ( x ), which is not Riemann integrable.

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2 INTRODUCTION TO LEBESGUE INTEGRATION 5 Example. Consider the function f n that connects the points (0 , 0) , (1 / 2 n, n ) , (1 /n, 0) , (1 , 0). These are Riemann integrable with Riemann integral 1 2 , but they converge to the zero function. Lebesgue’s Idea. Consider f : [ a, b ] ! R . Let the range of f be [ A, B ]. Let A = y 0 < y 1 < · · · < y n = B be a partition of the range. For each subinterval, look at E i = f - 1 [ y i - 1 , y i ] = { x : f ( x ) 2 [ y i - 1 , y i ] } . Then f P i y i χ E i . So take Z B A = “ lim X y i length( E i )” Lecture 3: May 8 We want to define a function m which measures the size of subsets of R and generalizes the notion of the length of an interval. Ideally, we want m : P ( R ) ! [0 , 1 ] = R + [ { 1 } to satisfy: (1) The measure of an interval should be the length of the interval. (2) m ( S 1 1 E j ) = P 1 1 m ( E j ) if the E j are disjoint. (This is called σ additivity.) (3) For t 2 R , let E + t = { x + t : x 2 E } . We want m ( E + t ) = m ( E ) 8 t 2 R 8 E R . (This is called translation invariance.) Notice that σ additivity implies that m is monotonic: If A B then m ( A ) m ( B ). Unfortunately, this is impossible, because of the Axiom of Choice. Theorem. Such a measure function, m , with the properties above, does not exist. Proof. Define a relation on R by x y if x - y 2 Q . This is an equivalence relation. Partition R into disjoint equivalence classes. Each class is of the form x 0 + Q . Each of these equivalence classes are dense in R , since they are just a translation of Q . In particular, each class intersects [0 , 1 2 ]. By the axiom of choice, we can pick one element in [0 , 1 2 ] from each equivalence class. Let E be the collection of these elements (one from each equivalence class). Now we claim that the sets E + x, x 2 Q are disjoint. Say ( E + x 1 ) \ ( E + x 2 ) 6 = ; . Then e 1 + x 1 = e 2 + x 2 for some e 1 , e 2 2 E . Then e 1 - e 2 = x 1 - x 2 2 Q , so e 1 e 2 . By the construction of E , e 1 = e 2 , and hence x 1 = x 2 . By σ -additivity, m S x 2 Q E + x = P x 2 Q m ( E + x ), since { E + x : x 2 Q } is a countable collection of disjoint sets. By translation invariance, we have m 0 @ [ x 2 Q E + x 1 A = X x 2 Q m ( E + x ) = X x 2 Q m ( E ) which equals 0 (if m ( E ) = 0) or 1 (if m ( E ) > 0). Now S x 2 Q E + x = R , so m S x 2 Q E + x = m ( R ) = 1 and hence, m ( E ) > 0.
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• Spring '12
• N.Spronk
• FN, Lebesgue measure, Lebesgue integration, Eastern Orthodox liturgical days, Lebesgue

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