# Lebesgue Integration

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 [Broken], 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

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Hurkyl
Staff Emeritus
Gold Member
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.

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.