Editing Dominated convergence theorem (section)

==Discussion of the assumptions==
The assumption that the sequence is dominated by some integrable ''g'' cannot be dispensed with. This may be seen as follows: define {{nowrap|''f<sub>n</sub>''(''x'') {{=}} ''n''}} for ''x'' in the [[interval (mathematics)|interval]] {{nowrap|(0, 1/''n'']}} and {{nowrap|''f''<sub>''n''</sub>(''x'') {{=}} 0}} otherwise. Any ''g'' which dominates the sequence must also dominate the pointwise [[supremum]] {{nowrap|''h'' {{=}} sup<sub>''n''</sub> ''f<sub>n</sub>''}}. Observe that
: <math>\int_0^1 h(x)\,dx \ge \int_{\frac{1}{m}}^1{h(x)\,dx} = \sum_{n=1}^{m-1} \int_{\left(\frac{1}{n+1},\frac{1}{n}\right]}{h(x)\,dx} \ge \sum_{n=1}^{m-1} \int_{\left(\frac{1}{n+1},\frac{1}{n}\right]}{n\,dx}=\sum_{n=1}^{m-1} \frac{1}{n+1} \to \infty \qquad \text{as }m\to\infty  </math>
by the divergence of the [[Harmonic series (mathematics)|harmonic series]]. Hence, the monotonicity of the Lebesgue integral tells us that there exists no integrable function which dominates the sequence on [0,1]. A direct calculation shows that integration and pointwise limit do not commute for this sequence:
: <math>\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,</math>
because the pointwise limit of the sequence is the [[zero function]]. Note that the sequence (''f<sub>n</sub>'') is not even [[uniformly integrable]], hence also the [[Vitali convergence theorem]] is not applicable.