Dominated convergence theorem

From Wikipedia, the free encyclopedia

In measure theory, a branch of mathematical analysis, Lebesgue's dominated convergence theorem provides sufficient conditions under which two limit processes commute, namely Lebesgue integration and pointwise convergence for a sequence of functions. This theorem shows the superiority of the Lebesgue integral over the Riemann integral for many theoretical purposes.

Contents

[edit] Statement of the theorem

Let f1, f2, f3, ... denote a sequence of real-valued measurable functions on a measure space (S,Σ,μ). Assume that the sequence converges pointwise and is dominated by some integrable function g. Then the pointwise limit is an integrable function and

\int_S\lim_{n\to\infty} f_n\,d\mu=\lim_{n\to\infty}\int_S f_n\,d\mu.

To say that the sequence is "dominated" by g means that

|f_n(x)| \le g(x)

for all natural numbers n and all points x in S. By integrable we mean

\int_S|g|\,d\mu<\infty.

The convergence of the sequence and domination by g can be relaxed to hold only μ-almost everywhere.

[edit] Proof of the theorem

Lebesgue's dominated convergence theorem is a special case of the Fatou–Lebesgue theorem. Below is a direct proof, using Fatou's lemma as the essential tool.

If f denotes the pointwise limit of the sequence, then f is also measurable and dominated by g, hence integrable. Furthermore,

|f-f_n|\le 2g

for all n and

\limsup_{n\to\infty}|f-f_n|=0.

By the reverse Fatou lemma,

\limsup_{n\to\infty}\int_S|f-f_n|\,d\mu \le\int_S\limsup_{n\to\infty}|f-f_n|\,d\mu=0.

Using linearity and monotonicity of the Lebesgue integral,

\biggl|\int_Sf\,d\mu-\int_Sf_n\,d\mu\biggr| =\biggl|\int_S(f-f_n)\,d\mu\biggr| \le\int_S|f-f_n|\,d\mu,

and the theorem follows.

[edit] Discussion of the assumptions

That the assumption that the sequence is dominated by some integrable g can not be dispensed with may be seen as follows: define fn(x) = n for x in the interval (0,1/n] and fn(x) = 0 otherwise. Any g which dominates the sequence must also dominate the pointwise supremum h = supn fn. 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}\frac1{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 [0,1]. 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.

[edit] Extensions

The theorem applies also to measurable functions with values in a Banach space, with the dominating function still being non-negative and integrable as above.

[edit] See also

[edit] References

  • R.G. Bartle, "The Elements of Integration and Lebesgue Measure", Wiley Interscience, 1995.
  • H.L. Royden, "Real Analysis", Prentice Hall, 1988.
  • D. Williams, "Probability with Martingales", Cambridge University Press, 1991, ISBN 0-521-40605-6