We introduced measure and Lebesgue Integral due to the deficiency of Riemann Integral. In general, there are at least three problematic issues about Riemann Integral:
Riemann integration does not handle functions with many discontinuities.
Riemann integration does not handle unbounded functions.
Riemann integration does not work well with limits.
Definition. (Riemann sum)
令 f:[a,b]→R 为一个有界实值函数,其中 a<b 为实数。在定义域 [a,b] 上的一个 partition,为一个有限集 P={a0,a1,a2,…,an},满足 a=a0<a1<a2<…<an=b。Partition P 从而定义了函数 f 的 lower Riemann sum 和 upper Riemann sum:
Then: L(P,f)=0 and U(P,f)=n∑i=1(ai−ai−1)=1 for any partition P, and hence the upper integral does not equal to the lower integral, since:
sup{L(P,f):P is a partition on [a,b]}=0inf{U(P,f):P is a partition on [a,b]}=1
Therefore we can conclude that f is not Riemann integrable, but since there are far more irrational number than rationals numbers, f should, in some sense, have integral 0. However, the Riemann integral of f is not defined.
Example 5.2(Problem with unbounded functions}
Define f:[0,1]→R by:
f(x)={1√xif 0<x≤10if x=0
For any partition P:={a0,a1,...,an} on [0, 1], for the first interval [a0,a1]=[0,a1], we always have:
supa0=0≤x≤a1f(x)⋅a1=∞
and then for any partition P of [0,1]:
U(P,f):=n∑i=1supai−1≤x≤aif(x)⋅(ai−ai−1)=∞
However, we should consider the area under the graph of f to be 2, not ∞, because lima→0+∫1af(x)dx=lima→0+(2−2√a)=2. Calculus courses deal with this example by defining ∫101√xdx=lima→0+∫1a1√xdx.That is, we can use the approach in Calculus to calculating Riemann integrals for these functions.
Example 5.3(Problem with unbounded functions)
Using the same approach above, and
f(x)=1√x+1√1−x
Then we would define ∫10f(x)dx to be:
∫10f(x)dx=lima→0+∫12af(x)dx+limb→1−∫b12f(x)dx
However, this technique (i.e. taking Riemann integrals over subdomains and then taking limits) can fail with more complicated functions, as shown in the following example.
Example 5.4(Problem with unbounded functions)
Let {rk}k∈Z+, be a sequence which includes each rational number in (0,1) exactly once. For k∈Z+, define fk:[0,1]→R by:
fk(x)={1√x−rk if x>rk0 if x≤rk
Define f:[0,1]→[0,∞] by:
f(x)=∞∑k=1fk(x)2k
For every non-empty open sub-interval (a,b)⊂[0,1], it must contains a rational number rh∈(a,b), then a<rh<b. Note that rh is also belongs to the sequence {rk}k∈Z+ because rh is a rational number in [0,1]. Hence for that rh, we have:
fh(x)={1√x−rh if x>rh0 if x≤rh
For x→r+h,fh(x)→∞, meaning fh(x) is unbounded. Thus, function f is unbounded on every its non-empty open sub-interval. Therefore, the Riemann integral of f is undefined on every its non-empty open sub-interval.
However, the area under the graph of each fk is less than 2:
∫10fk(x)dx=limt→1−,s→r+k∫ts1√x−rkdx=2√1−rk<2
Then the formula defining f:f(x)=∞∑k=1fk(x)2k shows that we should expect the area under the graph of f to be less than 2, instead of being undefined.
Example 5.5(Problem with point-wise limits)
Let {rk}k∈Z+, be a sequence which includes each rational number in [0,1] exactly once. For k∈Z+, define fk:[0,1]→R by:
fk(x)={1 if x∈{r1,...,rk}0 otherwise
Remark: Note that the difference between each function fk in this example and function f mentioned in Example 5.1 is that, fk is a function which takes 1 on finite points r1,...,rk, while f is a function which takes 1 on all rational numbers on [0,1].
For any k∈Z+, function fk is Riemann integrable: For lower integral, L(P,fk)=0 for any partition P on [0,1], since infai−1≤x≤aifk(x)=0 for all ai−1,ai∈P. Hence the lower integral of fk is 0.
For upper integral and the pair am−1,am such that rk∈[am−1,am], we have:
supam−1≤x≤amfk(x)⋅(am−am−1)=am−am−1,
and fk=0 on other intervals which do not include rk, thus inf(am−am−1)=0 by refining partition P. Hence the upper integral of fk is also 0. Because both lower and upper integral are 0, fk is Riemann integrable with ∫10fk(x)dx=0.
Define f:[0,1]→R by:
f(x)={1 if x is rational0 if x is irrational
Clearly limk→∞fk(x)=f(x) for each x∈[0,1].
However, f is not Riemann integrable (Example 5.1) even though f is the point-wise limit of a sequence of integrable functions bounded by 1.
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