Monotone convergence theorem

Template:Short description In the mathematical field of real analysis, the monotone convergence theorem is any of a number of related theorems proving the good convergence behaviour of monotonic sequences, i.e. sequences that are non-increasing, or non-decreasing. In its simplest form, it says that a non-decreasing bounded-above sequence of real numbers $a_1\le a_2\le a_3\le ...\le K$ converges to its smallest upper bound, its supremum. Likewise, a non-increasing bounded-below sequence converges to its largest lower bound, its infimum. In particular, infinite sums of non-negative numbers converge to the supremum of the partial sums if and only if the partial sums are bounded.

For sums of non-negative increasing sequences $0\le a_{i,1}\le a_{i,2}\le \cdots$, it says that taking the sum and the supremum can be interchanged.

In more advanced mathematics the monotone convergence theorem usually refers to a fundamental result in measure theory due to Lebesgue and Beppo Levi that says that for sequences of non-negative pointwise-increasing measurable functions $0\le f_1(x)\le f_2(x)\le \cdots$, taking the integral and the supremum can be interchanged with the result being finite if either one is finite.

Convergence of a monotone sequence of real numbers

Theorem: Let $(a_n{)}_{n\in \mathbb{\mathbb{N}}}$ be a monotone sequence of real numbers (either $a_n\le a_{n+1}$ for all $n$ or $a_n\ge a_{n+1}$ for all $n$). Then the following are equivalent:

1. $(a_n)$ has a finite limit in $\mathbb{ℝ}$.
2. $(a_n)$ is bounded.

Moreover, if $(a_n)$ is nondecreasing, then $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\underset{n}{\sup }{a}_n$; if $(a_n)$ is nonincreasing, then $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\underset{n}{\inf }{a}_n$.

Proof

(1 ⇒ 2) Suppose $(a_n)\to L\in \mathbb{ℝ}$. By the $\epsilon$-definition of limit, there exists $N$ such that $|a_n-L|<1$ for all $n\ge N$, hence $|a_n|\le |L|+1$ for $n\ge N$. Let $M=\max \{|a_1|,\dots ,|a_{N-1}|,|L|+1\}$. Then $|a_n|\le M$ for all $n$, so $(a_n)$ is bounded.

(2 ⇒ 1) Suppose $(a_n)$ is bounded and monotone.

• If $(a_n)$ is nondecreasing and bounded above, set $c=\underset{n}{\sup }{a}_n$. For any $\epsilon >0$, there exists $N$ with $c-\epsilon <a_N\le c$; otherwise $c-\epsilon$ would be a smaller upper bound than $c$. For $n\ge N$, monotonicity gives $a_N\le a_n\le c$, hence $0\le c-a_n\le c-a_N<\epsilon$. Thus $a_n\to c=\underset{n}{\sup }{a}_n$.
• If $(a_n)$ is nonincreasing and bounded below, either repeat the argument with $c=\underset{n}{\inf }{a}_n$, or apply the previous case to $(-a_n)$ to obtain $a_n\to \underset{n}{\inf }{a}_n$.

This proves the equivalence.

Remark

The implication "bounded and monotone ⇒ convergent" may fail over $\mathbb{\mathbb{Q}}$ because the supremum/infimum of a rational sequence need not be rational. For example, $a_n=\lfloor 10^n\sqrt{2}\rfloor /10^n$ is nondecreasing and bounded above by $\sqrt{2}$, but has no limit in $\mathbb{\mathbb{Q}}$ (its real limit is $\sqrt{2}$).

Convergence of a monotone series

There is a variant of the proposition above where we allow unbounded sequences in the extended real numbers, the real numbers with $\mathrm{\infty }$ and $-\mathrm{\infty }$ added.

$\overline{\mathbb{ℝ}}=\mathbb{ℝ}\cup \{-\mathrm{\infty },\mathrm{\infty }\}$

In the extended real numbers every set has a supremum (resp. infimum) which of course may be $\mathrm{\infty }$ (resp. $-\mathrm{\infty }$) if the set is unbounded. An important use of the extended reals is that any set of non negative numbers $a_i\ge 0,i\in I$ has a well defined summation order independent sum

$\sum _{i\in I}{a}_i=\underset{J\subset I,|J|<\mathrm{\infty }}{\sup }\sum _{j\in J}{a}_j\in {\overline{\mathbb{ℝ}}}_{\ge 0}$

where ${\overline{\mathbb{ℝ}}}_{\ge 0}=[0,\mathrm{\infty }]\subset \overline{\mathbb{ℝ}}$ are the upper extended non negative real numbers. For a series of non negative numbers

$\sum _{i=1}^{\mathrm{\infty }}{a}_i=\underset{k\to \mathrm{\infty }}{\lim }\sum _{i=1}^k{a}_i=\underset{k}{\sup }\sum _{i=1}^k{a}_i=\underset{J\subset \mathbb{\mathbb{N}},|J|<\mathrm{\infty }}{\sup }\sum _{j\in J}{a}_j=\sum _{i\in \mathbb{\mathbb{N}}}{a}_i,$

so this sum coincides with the sum of a series if both are defined. In particular the sum of a series of non negative numbers does not depend on the order of summation.

Monotone convergence of non negative sums

Let $a_{i,k}\ge 0$ be a sequence of non-negative real numbers indexed by natural numbers $i$ and $k$. Suppose that $a_{i,k}\le a_{i,k+1}$ for all $i,k$. ThenTemplate:Rp

$\underset{k}{\sup }\sum _i{a}_{i,k}=\sum _i\underset{k}{\sup }{a}_{i,k}\in {\overline{\mathbb{ℝ}}}_{\ge 0}.$

Proof

Since $a_{i,k}\le \underset{k}{\sup }{a}_{i,k}$ we have $\sum _i{a}_{i,k}\le \sum _i\underset{k}{\sup }{a}_{i,k}$ so $\underset{k}{\sup }\sum _i{a}_{i,k}\le \sum _i\underset{k}{\sup }{a}_{i,k}$.

Conversely, we can interchange sup and sum for finite sums by reverting to the limit definition, so $\sum _{i=1}^N\underset{k}{\sup }{a}_{i,k}=\underset{k}{\sup }\sum _{i=1}^N{a}_{i,k}\le \underset{k}{\sup }\sum _{i=1}^{\mathrm{\infty }}{a}_{i,k}$ hence $\sum _{i=1}^{\mathrm{\infty }}\underset{k}{\sup }{a}_{i,k}\le \underset{k}{\sup }\sum _{i=1}^{\mathrm{\infty }}{a}_{i,k}$.

Examples

Matrices

The theorem states that if you have an infinite matrix of non-negative real numbers $a_{i,k}\ge 0$ such that the rows are weakly increasing and each is bounded $a_{i,k}\le K_i$ where the bounds are summable $\sum _i{K}_i<\mathrm{\infty }$ then, for each column, the non decreasing column sums $\sum _i{a}_{i,k}\le \sum K_i$ are bounded hence convergent, and the limit of the column sums is equal to the sum of the "limit column" $\underset{k}{\sup }{a}_{i,k}$ which element wise is the supremum over the row.

e

Consider the expansion

${(1+\frac{1}{k})}^k=\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})\frac{1}{k^i}$

Now set

$a_{i,k}=(\genfrac{}{}{0ex}{}{k}{i})\frac{1}{k^i}=\frac{1}{i!}\cdot \frac{k}{k}\cdot \frac{k-1}{k}\cdot \cdots \frac{k-i+1}{k}$

for $i\le k$ and $a_{i,k}=0$ for $i>k$, then $0\le a_{i,k}\le a_{i,k+1}$ with $\underset{k}{\sup }{a}_{i,k}=\frac{1}{i!}<\mathrm{\infty }$ and

${(1+\frac{1}{k})}^k=\sum _{i=0}^{\mathrm{\infty }}{a}_{i,k}$.

The right hand side is a non decreasing sequence in $k$, therefore

$\underset{k\to \mathrm{\infty }}{\lim }{(1+\frac{1}{k})}^k=\underset{k}{\sup }\sum _{i=0}^{\mathrm{\infty }}{a}_{i,k}=\sum _{i=0}^{\mathrm{\infty }}\underset{k}{\sup }{a}_{i,k}=\sum _{i=0}^{\mathrm{\infty }}\frac{1}{i!}=e$.

Monotone convergence for non-negative measurable functions (Beppo Levi)

The following result extends the monotone convergence of non-negative series to the measure-theoretic setting. It is a cornerstone of measure and integration theory; Fatou's lemma and the dominated convergence theorem follow as direct consequences. It is due to Beppo Levi, who in 1906 proved a slight generalization of an earlier result by Henri Lebesgue.

Let ${{ℬ}}_{{\overline{\mathbb{ℝ}}}_{\ge 0}}$ denote the Borel $\sigma$-algebra on the extended half-line $[0,+\mathrm{\infty }]$ (so $\{+\mathrm{\infty }\}\in {{ℬ}}_{{\overline{\mathbb{ℝ}}}_{\ge 0}}$).

Theorem (Monotone convergence for non-negative measurable functions)

Let $(\mathrm{\Omega },\mathrm{\Sigma },\mu )$ be a measure space and $X\in \mathrm{\Sigma }$. If $\{f_k{\}}_{k\ge 1}$ is a sequence of non-negative $(\mathrm{\Sigma },{{ℬ}}_{{\overline{\mathbb{ℝ}}}_{\ge 0}})$-measurable functions on $X$ such that $0\le f_1(x)\le f_2(x)\le \cdots \text{for all }x\in X,$ then the pointwise supremum $f:=\underset{k}{\sup }{f}_k$ is measurable and $\underset{X}{\int }fd\mu =\underset{k\to \mathrm{\infty }}{\lim }\underset{X}{\int }{f}_{k}d\mu =\underset{k}{\sup }\underset{X}{\int }{f}_{k}d\mu .$

Proof

Let $f=\underset{k}{\sup }{f}_k$. Measurability of $f$ follows since pointwise limits/suprema of measurable functions are measurable.

Upper bound. By monotonicity of the integral, $f_k\le f$ implies $\underset{k}{\mathrm{lim\; sup}}\underset{X}{\int }{f}_{k}d\mu \le \underset{X}{\int }fd\mu .$

Lower bound. Fix a non-negative simple function $s<f$. Set $A_k=\{x\in X:s(x)\le f_k(x)\}.$ Then $A_k\uparrow X$ because $f_k\uparrow f\ge s$. For the set function ${\nu }_s(A):=\underset{A}{\int }sd\mu ,$ we have ${\nu }_s$ is a measure (write $s=\sum _i{c}_i{\mathbf{𝟏}}_{E_i}$ and note ${\nu }_s(A)=\sum _i{c}_i\mu (A\cap E_i)$), hence by continuity from below, $\underset{X}{\int }sd\mu =\underset{k\to \mathrm{\infty }}{\lim }\underset{A_k}{\int }sd\mu .$ On each $A_k$ we have $s\le f_k$, so $\underset{A_k}{\int }sd\mu \le \underset{X}{\int }{f}_{k}d\mu .$ Taking limits gives $\underset{X}{\int }sd\mu \le \underset{k}{\mathrm{lim\; inf}}\underset{X}{\int }{f}_{k}d\mu$. Finally, take the supremum over all simple $s<f$ (which equals $\underset{X}{\int }fd\mu$ by definition of the Lebesgue integral) to obtain $\underset{X}{\int }fd\mu \le \underset{k}{\mathrm{lim\; inf}}\underset{X}{\int }{f}_{k}d\mu .$

Combining the two bounds yields $\underset{X}{\int }fd\mu =\underset{k\to \mathrm{\infty }}{\lim }\underset{X}{\int }{f}_{k}d\mu =\underset{k}{\sup }\underset{X}{\int }{f}_{k}d\mu .◻$

Remarks

1. (Finiteness.) The quantities may be finite or infinite; the left-hand side is finite iff the right-hand side is.
2. (Pointwise and integral limits.) Under the hypotheses,
   • $\underset{k\to \mathrm{\infty }}{\lim }{f}_k(x)=\underset{k}{\sup }{f}_k(x)=\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}{f}_k(x)=\underset{k\to \mathrm{\infty }}{\mathrm{lim\; inf}}{f}_k(x)$ for all $x$;
   • by monotonicity of the integral, $\underset{k\to \mathrm{\infty }}{\lim }\underset{X}{\int }{f}_{k}d\mu =\underset{k}{\sup }\underset{X}{\int }{f}_{k}d\mu =\underset{k\to \mathrm{\infty }}{\mathrm{lim\; inf}}\underset{X}{\int }{f}_{k}d\mu =\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\underset{X}{\int }{f}_{k}d\mu .$ Equivalently, $\underset{k\to \mathrm{\infty }}{\lim }\underset{X}{\int }{f}_{k}d\mu =\underset{X}{\int }\underset{k\to \mathrm{\infty }}{\lim }{f}_{k}d\mu ,$ with the understanding that the limits may be $+\mathrm{\infty }$.
3. (Almost-everywhere version.) If the monotonicity holds $\mu$-almost everywhere, then redefining the limit function arbitrarily on a null set preserves measurability and leaves all integrals unchanged. Hence the theorem still holds.
4. (Foundational role.) The proof uses only: (i) monotonicity of the integral for non-negative functions; (ii) that $A\mapsto \underset{A}{\int }sd\mu$ is a measure for simple $s$; and (iii) continuity from below of measures. Thus the lemma can be used to derive further basic properties (e.g. linearity) of the Lebesgue integral.
5. (Relaxing the monotonicity assumption.) Under similar hypotheses, one can relax monotonicity. Let $(\mathrm{\Omega },\mathrm{\Sigma },\mu )$ be a measure space, $X\in \mathrm{\Sigma }$, and let $\{f_k{\}}_{k\ge 1}$ be non-negative measurable functions on $X$ such that $f_k(x)\to f(x)$ for a.e. $x$ and $f_k\le f$ a.e. for all $k$. Then $f$ is measurable, the limit $\underset{k\to \mathrm{\infty }}{\lim }\underset{X}{\int }{f}_{k}d\mu$ exists, and $\underset{k\to \mathrm{\infty }}{\lim }\underset{X}{\int }{f}_{k}d\mu =\underset{X}{\int }fd\mu .$

Proof based on Fatou's lemma

The proof can also be based on Fatou's lemma instead of a direct proof as above, because Fatou's lemma can be proved independent of the monotone convergence theorem. However the monotone convergence theorem is in some ways more primitive than Fatou's lemma. It easily follows from the monotone convergence theorem and proof of Fatou's lemma is similar and arguably slightly less natural than the proof above.

As before, measurability follows from the fact that $f=\underset{k}{\sup }{f}_k=\underset{k\to \mathrm{\infty }}{\lim }{f}_k=\underset{k\to \mathrm{\infty }}{\mathrm{lim\; inf}}{f}_k$ almost everywhere. The interchange of limits and integrals is then an easy consequence of Fatou's lemma. One has $$\underset{X}{\int }fd\mu =\underset{X}{\int }\underset{k}{\mathrm{lim\; inf}}{f}_{k}d\mu \le \mathrm{lim\; inf}\underset{X}{\int }{f}_{k}d\mu$$ by Fatou's lemma, and then, since $\int f_{k}d\mu \le \int f_{k+1}d\mu \le \int fd\mu$ (monotonicity), $$\mathrm{lim\; inf}\underset{X}{\int }{f}_{k}d\mu \le \underset{k}{\mathrm{lim\; sup}}\underset{X}{\int }{f}_{k}d\mu =\underset{k}{\sup }\underset{X}{\int }{f}_{k}d\mu \le \underset{X}{\int }fd\mu .$$ Therefore $$\underset{X}{\int }fd\mu =\underset{k\to \mathrm{\infty }}{\mathrm{lim\; inf}}\underset{X}{\int }{f}_{k}d\mu =\underset{k\to \mathrm{\infty }}{\mathrm{lim\; sup}}\underset{X}{\int }{f}_{k}d\mu =\underset{k\to \mathrm{\infty }}{\lim }\underset{X}{\int }{f}_{k}d\mu =\underset{k}{\sup }\underset{X}{\int }{f}_{k}d\mu .$$

See also

• Infinite series
• Dominated convergence theorem