Implications of Lebesgue Integration for Bounded Functions

[wayneckm]

Hello all,


I am wondering the implication between almost everywhere bounded function and Lebesgue integrable.

In the theory of Lebesgue integration, if a non-negative function $$f$$ is bounded a.e., then it should be Lebesgue integrable, i.e. $$\int f d\mu < \infty$$ because we do not take into account the unboundedness of $$f$$ in a null set when approximate by sequence of simple function, am I correct? So this means a.e. boundedness implies Lebesgue integrable?

And seems there is a counterexample on the reverse implication, http://planetmath.org/encyclopedia/AnIntegrableFunctionWhichDoesNotTendToZero.html , so that means Lebesgue integrable does not imply bounded a.e.

So is this because in finding the Lebesgue integral, it is indeed an infinite series of products, which is $$\sum s_{n} \cdot \mu(A_{n})$$, so as long as the increase in $$s_{n}$$ is not faster than the decrease in $$\mu(A_{n})$$, it is possible to have a finite value of this infinite sum? So in this way we may end up with a non-bounded a.e. function but Lebesgue integrable?


Wayne

 

[Hurkyl]

One standard example of a non-Lebesgue integrable function is the characteristic function of a non-measurable set.

While the Lebesgue integral (and the Riemann integral!) are limits of finite sums, neither is an infinite sum. (In the usual formulations, anyways)

 

[mathman]

So this means a.e. boundedness implies Lebesgue integrable?

This assumes that the total measure of the space is finite.

 

[wayneckm]

Thanks so much.

So, in conclusion, relationship between a.e. boundedness and Lebesgue integrable is not definitive in general, right?

Regarding the characteristic function of a non-measurable set, it is then a non-measurable function, hence, its Lebesgue integral is not well-defined? Or is there any reference about this?

Thanks.