College Math Teaching

February 3, 2011

Why A Bounded Condition is Necessary for Intergral Convergence Theorems

Filed under: advanced mathematics, analysis, calculus, integrable function, integrals, Lebesgue Integration — collegemathteaching @ 1:07 am

One of the reasons that the Lebesgue theory is an improvement on the Riemann theory is that we have better convergence properties for integrals. For example, we could define the following sequence of functions: f _{k}(x)=\left\{ \begin{array}{c}1,x\notin {1/2, 1/4, 3/4. ...1/2^k, 3/2^k,...(2^{k-1} -1)/2^k} \\ 0,x\in {1/2, 1/4, 3/4. ...1/2^k, 3/2^k,...(2^{k-1} -1)/2^k} \end{array}\right. Then for each k , \int_{0}^{1} f_{k}(x)dx =1 for all k and f_{k}(x) \to f(x) point wise and f does not have a Riemann integral. It does have a Lebesgue integral which is 1.

In fact, there are many convergence theorems, one of which is the Bounded Convergence Theorem:
If f_{k} is a sequence of measurable functions on a finite measure set S and the functions f_{k} are uniformly bounded on S and f_{n} \to f pointwise on S then lim \int_{S} f_{k} = \int_{S} lim f_{k} = \int_{S} f .

We’ll take on this convergence theorem a little later (in another post). But why do we need a condition of the “uniformly bounded” type?
We’ll present an elementary but fun counter example to the theorem if the “uniformly bounded” condition is dropped.

Well start our construction by considering the sequence {1/2, 1/3, ....1/k,...} . We then note that 1/n and 1/n+1 are 1/(n)(n+1) units apart hence we can form disjoint segments ( 1/n - 1/(3(n)(n+1)), 1/n + 1/(3(n)(n+1))) which can be used as the base for disjoint isosceles triangles whose upper vertex is at coordinate (1/n, (3n)(n+1)) . Each of these triangles enclose an area of 1.
Now we can define a sequence of functions f_{k} by letting f_{k} = 0 off the base of the triangle centered at 1/k and letting the graph of f_{k} follow the upper two edges of the triangle. Then for all k we have \int_{0}^{1} f_{k}(x)dx = 1 . Hence lim \int_{0}^{1} f_{k} (x)dx = 1 . Also note that f_{k} \to f where f(x) = 0 for all x \in [0,1] . This isn’t hard to see; first note that f_{k}(0) = 0 for all k and for t > 0 f_{m} (t) = 0 for all m where (3m + 4)/(3m^2 +3m) < t . Of course, \int_{0}^{1} f(x)dx = 0 . Each f_{k} is bounded but the sequence is NOT uniformly bounded as the peaks of the triangles get higher and higher.

Note: I made a mistake when I first posted this; of course we don’t NEED “uniform boundedness” but we need either a boundedness condition or some condition that ensures that the measure of the unbounded parts is zero.

Example: order the rationals between 0 and 1 by q_{i} and let each rational be written as p_{i}/d_{i} in lowest terms.
Then define f_{k}(x) to be d_{i} for x \in {q_{1},...,q_{k}} and 1 otherwise. Then the f_{k} are NOT uniformly bounded as sequence but \int_{0}^{1}f_{k} = 1 for all k and \int_{0}^{1}f =1 (where f_{k}\to f ).
But this is ok as f differs from the constant function y = 1 on a set of measure zero.


1 Comment »

  1. […] and perhaps play with MATLAB and possibly write up another post for the mathematics blog. I had fun with this one, which was about why we need convergence theorems for integrals (that is, the limit of the […]

    Pingback by 3 February 2011: the day after and the big freeze « blueollie — February 3, 2011 @ 12:39 pm

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