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

In the theory of Lebesgue integration, if a non-negative function [tex]f[/tex] is bounded a.e., then it should be Lebesgue integrable, i.e. [tex]\int f d\mu < \infty[/tex] because we do not take into account the unboundedness of [tex]f[/tex] 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 [tex] \sum s_{n} \cdot \mu(A_{n})[/tex], so as long as the increase in [tex] s_{n} [/tex] is not faster than the decrease in [tex] \mu(A_{n}) [/tex], 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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# Lebesgue Integration

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