Discussion of The Assumptions
The assumption that the sequence is dominated by some integrable g can not be dispensed with. This may be seen as follows: define ƒn(x) = n for x in the interval (0, 1/n] and ƒn(x) = 0 otherwise. Any g which dominates the sequence must also dominate the pointwise supremum h = supn ƒn. Observe that
![\int_0^1 h(x)\,dx \ge \int_{1/m}^1{h(x)\,dx} = \sum_{n=1}^{m-1} \int_{\left(\frac1{n+1},\frac1n\right]}{n\,dx} = \sum_{n=1}^{m-1} \frac{1}{n+1} \to \infty \quad \text{as }m\to\infty](http://upload.wikimedia.org/math/2/a/d/2addf2eec53134138612631fc7b9b2a4.png)
by the divergence of the harmonic series. Hence, the monotonicity of the Lebesgue integral tells us that there exists no integrable function which dominates the sequence on . A direct calculation shows that integration and pointwise limit do not commute for this sequence:
because the pointwise limit of the sequence is the zero function. Note that the sequence {ƒn} is not even uniformly integrable, hence also the Vitali convergence theorem is not applicable.
Read more about this topic: Dominated Convergence Theorem
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