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
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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![\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)
