# 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.