Dominated Convergence Theorem - Discussion of The Assumptions

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

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:

\int_0^1 \lim_{n\to\infty} f_n(x)\,dx = 0 \neq 1 = \lim_{n\to\infty}\int_0^1 f_n(x)\,dx,

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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