Dominated convergence theorem

Proof

Without loss of generality, one can assume that f is real, because one can split f into its real and imaginary parts (remember that a sequence of complex numbers converges if and only if both its real and imaginary counterparts converge) and apply the triangle inequality at the end.

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

Since f is the pointwise limit of the sequence (f_n) of measurable functions that are dominated by g, it is also measurable and dominated by g, hence it is integrable. Furthermore, (these will be needed later),

${\displaystyle |f-f_n|\le |f|+|f_n|\le 2g}$

for all n and

${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}|f-f_n|=0}$

where ${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}}$ is the limit superior.

The second of these is trivially true (by the very definition of f). Using linearity and monotonicity of the Lebesgue integral,

${\displaystyle \left|{\int }_{S}fd\mu -{\int }_S{f}_{n}d\mu \right|=\left|{\int }_S(f-f_n)d\mu \right|\le {\int }_S|f-f_n|d\mu .}$

By the reverse Fatou lemma (it is here that we use the fact that |f−f_n| is bounded above by an integrable function)

${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}{\int }_S|f-f_n|d\mu \le {\int }_S\underset{n\to \mathrm{\infty }}{lim sup}|f-f_n|d\mu =0,}$

which implies that the limit exists and vanishes i.e.

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_S|f-f_n|d\mu =0.}$

Finally, since

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\int }_{S}fd\mu -{\int }_S{f}_{n}d\mu \right|\le \underset{n\to \mathrm{\infty }}{lim}{\int }_S|f-f_n|d\mu =0.}$

we have that

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_S{f}_{n}d\mu ={\int }_{S}fd\mu .}$

The theorem now follows.

If the assumptions hold only μ-almost everywhere, then there exists a μ-null set N ∈ Σ such that the functions f_n 1_{S \ N} satisfy the assumptions everywhere on S. Then the function f(x) defined as the pointwise limit of f_n(x) for x ∈ S \ N and by f(x) = 0 for x ∈ N, is measurable and is the pointwise limit of this modified function sequence. The values of these integrals are not influenced by these changes to the integrands on this μ-null set N, so the theorem continues to hold.

DCT holds even if f_n converges to f in measure (finite measure) and the dominating function is non-negative almost everywhere.

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 f_n(x) = n for x in the interval (0, 1/n] and f_n(x) = 0 otherwise. Any g which dominates the sequence must also dominate the pointwise supremum h = sup_n f_n. Observe that

${\displaystyle {\int }_0^{1}h(x)dx\ge {\int }_{\frac{1}{m}}^{1}h(x)dx=\sum _{n=1}^{m-1}{\int }_{\left(\frac{1}{n+1},\frac{1}{n}\right]}h(x)dx\ge \sum _{n=1}^{m-1}{\int }_{\left(\frac{1}{n+1},\frac{1}{n}\right]}ndx=\sum _{n=1}^{m-1}\frac{1}{n+1}\to \mathrm{\infty }\text{as }m\to \mathrm{\infty }}$

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:

${\displaystyle {\int }_0^1\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)dx=0\ne 1=\underset{n\to \mathrm{\infty }}{lim}{\int }_0^1{f}_n(x)dx,}$

because the pointwise limit of the sequence is the zero function. Note that the sequence (f_n) is not even uniformly integrable, hence also the Vitali convergence theorem is not applicable.

Bounded convergence theorem

One corollary to the dominated convergence theorem is the bounded convergence theorem, which states that if (f_n) is a sequence of uniformly bounded complex-valued measurable functions which converges pointwise on a bounded measure space (S, Σ, μ) (i.e. one in which μ(S) is finite) to a function f, then the limit f is an integrable function and

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_S{f}_{n}d\mu ={\int }_{S}fd\mu .}$

Remark: The pointwise convergence and uniform boundedness of the sequence can be relaxed to hold only μ-almost everywhere, provided the measure space (S, Σ, μ) is complete or f is chosen as a measurable function which agrees μ-almost everywhere with the μ-almost everywhere existing pointwise limit.

Proof

Since the sequence is uniformly bounded, there is a real number M such that |f_n(x)| ≤ M for all x ∈ S and for all n. Define g(x) = M for all x ∈ S. Then the sequence is dominated by g. Furthermore, g is integrable since it is a constant function on a set of finite measure. Therefore, the result follows from the dominated convergence theorem.

If the assumptions hold only μ-almost everywhere, then there exists a μ-null set N ∈ Σ such that the functions f_n1_S\N satisfy the assumptions everywhere on S.

Dominated convergence in Lp-spaces (corollary)

Let ${\displaystyle (\mathrm{\Omega },\mathcal{A},\mu )}$ be a measure space, ${\displaystyle 1\le p<\mathrm{\infty }}$ a real number and ${\displaystyle (f_n)}$ a sequence of ${\displaystyle \mathcal{A}}$-measurable functions ${\displaystyle f_n:\mathrm{\Omega }\to \mathbb{C}\cup \{\mathrm{\infty }\}}$.

Assume the sequence ${\displaystyle (f_n)}$ converges ${\displaystyle \mu }$-almost everywhere to an ${\displaystyle \mathcal{A}}$-measurable function ${\displaystyle f}$, and is dominated by a ${\displaystyle g\in L^p}$ (cf. Lp space), i.e., for every natural number ${\displaystyle n}$ we have: ${\displaystyle |f_n|\le g}$, μ-almost everywhere.

Then all ${\displaystyle f_n}$ as well as ${\displaystyle f}$ are in ${\displaystyle L^p}$ and the sequence ${\displaystyle (f_n)}$ converges to ${\displaystyle f}$ in the sense of ${\displaystyle L^p}$, i.e.:

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\Vert f_n-f{\Vert }_p=\underset{n\to \mathrm{\infty }}{lim}{\left({\int }_{\mathrm{\Omega }}|f_n-f{|}^{p}d\mu \right)}^{\frac{1}{p}}=0.}$

Idea of the proof: Apply the original theorem to the function sequence ${\displaystyle h_n=|f_n-f{|}^p}$ with the dominating function ${\displaystyle (2g{)}^p}$.

Extensions

The dominated convergence theorem applies also to measurable functions with values in a Banach space, with the dominating function still being non-negative and integrable as above. The assumption of convergence almost everywhere can be weakened to require only convergence in measure.

The dominated convergence theorem applies also to conditional expectations.