Monotone convergence theorem

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 ${\displaystyle a_{1}\leq a_{2}\leq a_{3}\leq ...\leq K}$ 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 ${\displaystyle 0\leq a_{i,1}\leq a_{i,2}\leq \cdots }$, 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 ${\displaystyle 0\leq f_{1}(x)\leq f_{2}(x)\leq \cdots }$, taking the integral and the supremum can be interchanged with the result being finite if either one is finite.

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

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

Moreover, if ${\displaystyle (a_{n})}$ is nondecreasing, then ${\displaystyle \lim _{n\to \infty }a_{n}=\sup _{n}a_{n}}$; if ${\displaystyle (a_{n})}$ is nonincreasing, then ${\displaystyle \lim _{n\to \infty }a_{n}=\inf _{n}a_{n}}$.^[1]

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

(2 ⇒ 1) Suppose ${\displaystyle (a_{n})}$ is bounded and monotone.

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

This proves the equivalence.

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

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

${\displaystyle {\bar {\mathbb {R} }}=\mathbb {R} \cup \{-\infty ,\infty \}}$

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

${\displaystyle \sum _{i\in I}a_{i}=\sup _{J\subset I,\ |J|<\infty }\sum _{j\in J}a_{j}\in {\bar {\mathbb {R} }}_{\geq 0}}$

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

${\displaystyle \sum _{i=1}^{\infty }a_{i}=\lim _{k\to \infty }\sum _{i=1}^{k}a_{i}=\sup _{k}\sum _{i=1}^{k}a_{i}=\sup _{J\subset \mathbb {N} ,|J|<\infty }\sum _{j\in J}a_{j}=\sum _{i\in \mathbb {N} }a_{i},}$

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.

Let ${\displaystyle a_{i,k}\geq 0}$ be a sequence of non-negative real numbers indexed by natural numbers ${\displaystyle i}$ and ${\displaystyle k}$. Suppose that ${\displaystyle a_{i,k}\leq a_{i,k+1}}$ for all ${\displaystyle i,k}$. Then^[2]: 168

${\displaystyle \sup _{k}\sum _{i}a_{i,k}=\sum _{i}\sup _{k}a_{i,k}\in {\bar {\mathbb {R} }}_{\geq 0}.}$

Since ${\displaystyle a_{i,k}\leq \sup _{k}a_{i,k}}$ we have ${\displaystyle \sum _{i}a_{i,k}\leq \sum _{i}\sup _{k}a_{i,k}}$ so ${\displaystyle \sup _{k}\sum _{i}a_{i,k}\leq \sum _{i}\sup _{k}a_{i,k}}$.

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

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

Consider the expansion

${\displaystyle \left(1+{\frac {1}{k}}\right)^{k}=\sum _{i=0}^{k}{\binom {k}{i}}{\frac {1}{k^{i}}}}$

Now set

${\displaystyle a_{i,k}={\binom {k}{i}}{\frac {1}{k^{i}}}={\frac {1}{i!}}\cdot {\frac {k}{k}}\cdot {\frac {k-1}{k}}\cdot \cdots {\frac {k-i+1}{k}}}$

for ${\displaystyle i\leq k}$ and ${\displaystyle a_{i,k}=0}$ for ${\displaystyle i>k}$, then ${\displaystyle 0\leq a_{i,k}\leq a_{i,k+1}}$ with ${\displaystyle \sup _{k}a_{i,k}={\frac {1}{i!}}<\infty }$ and

${\displaystyle \left(1+{\frac {1}{k}}\right)^{k}=\sum _{i=0}^{\infty }a_{i,k}}$.

The right hand side is a non decreasing sequence in ${\displaystyle k}$, therefore

${\displaystyle \lim _{k\to \infty }\left(1+{\frac {1}{k}}\right)^{k}=\sup _{k}\sum _{i=0}^{\infty }a_{i,k}=\sum _{i=0}^{\infty }\sup _{k}a_{i,k}=\sum _{i=0}^{\infty }{\frac {1}{i!}}=e}$.

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.^[3][4]

Let ${\displaystyle \operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}}}$ denote the Borel ${\displaystyle \sigma }$-algebra on the extended half-line ${\displaystyle [0,+\infty ]}$ (so ${\displaystyle \{+\infty \}\in \operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}}}$).

Let ${\displaystyle (\Omega ,\Sigma ,\mu )}$ be a measure space and ${\displaystyle X\in \Sigma }$. If ${\displaystyle \{f_{k}\}_{k\geq 1}}$ is a sequence of non-negative ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})}$-measurable functions on ${\displaystyle X}$ such that ${\displaystyle 0\leq f_{1}(x)\leq f_{2}(x)\leq \cdots \quad {\text{for all }}x\in X,}$ then the pointwise supremum ${\displaystyle f:=\sup _{k}f_{k}}$ is measurable and ${\displaystyle \int _{X}f\,d\mu \;=\;\lim _{k\to \infty }\int _{X}f_{k}\,d\mu \;=\;\sup _{k}\int _{X}f_{k}\,d\mu .}$

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

Upper bound. By monotonicity of the integral, ${\displaystyle f_{k}\leq f}$ implies ${\displaystyle \limsup _{k}\int _{X}f_{k}\,d\mu \;\leq \;\int _{X}f\,d\mu .}$

Lower bound. Fix a non-negative simple function ${\displaystyle s\leq f}$. Set ${\displaystyle A_{k}=\{x\in X:\;s(x)\leq f_{k}(x)\}.}$ Then ${\displaystyle A_{k}\uparrow X}$ because ${\displaystyle f_{k}\uparrow f\geq s}$. For the set function ${\displaystyle \nu _{s}(A):=\int _{A}s\,d\mu ,}$ we have ${\displaystyle \nu _{s}}$ is a measure (write ${\displaystyle s=\sum _{i}c_{i}\mathbf {1} _{E_{i}}}$ and note ${\displaystyle \nu _{s}(A)=\sum _{i}c_{i}\,\mu (A\cap E_{i})}$), hence by continuity from below, ${\displaystyle \int _{X}s\,d\mu \;=\;\lim _{k\to \infty }\int _{A_{k}}s\,d\mu .}$ On each ${\displaystyle A_{k}}$ we have ${\displaystyle s\leq f_{k}}$, so ${\displaystyle \int _{A_{k}}s\,d\mu \;\leq \;\int _{X}f_{k}\,d\mu .}$ Taking limits gives ${\displaystyle \int _{X}s\,d\mu \leq \liminf _{k}\int _{X}f_{k}\,d\mu }$. Finally, take the supremum over all simple ${\displaystyle s\leq f}$ (which equals ${\displaystyle \int _{X}f\,d\mu }$ by definition of the Lebesgue integral) to obtain ${\displaystyle \int _{X}f\,d\mu \;\leq \;\liminf _{k}\int _{X}f_{k}\,d\mu .}$

Combining the two bounds yields ${\displaystyle \int _{X}f\,d\mu \;=\;\lim _{k\to \infty }\int _{X}f_{k}\,d\mu \;=\;\sup _{k}\int _{X}f_{k}\,d\mu .\square }$

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,
   • ${\displaystyle \displaystyle \lim _{k\to \infty }f_{k}(x)=\sup _{k}f_{k}(x)=\limsup _{k\to \infty }f_{k}(x)=\liminf _{k\to \infty }f_{k}(x)}$ for all ${\displaystyle x}$;
   • by monotonicity of the integral, ${\displaystyle \displaystyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu =\sup _{k}\int _{X}f_{k}\,d\mu =\liminf _{k\to \infty }\int _{X}f_{k}\,d\mu =\limsup _{k\to \infty }\int _{X}f_{k}\,d\mu .}$ Equivalently, ${\displaystyle \displaystyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu =\int _{X}\lim _{k\to \infty }f_{k}\,d\mu ,}$ with the understanding that the limits may be ${\displaystyle +\infty }$.
3. (Almost-everywhere version.) If the monotonicity holds ${\displaystyle \mu }$-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 ${\displaystyle A\mapsto \int _{A}s\,d\mu }$ is a measure for simple ${\displaystyle s}$; 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.^[5] Let ${\displaystyle (\Omega ,\Sigma ,\mu )}$ be a measure space, ${\displaystyle X\in \Sigma }$, and let ${\displaystyle \{f_{k}\}_{k\geq 1}}$ be non-negative measurable functions on ${\displaystyle X}$ such that ${\displaystyle f_{k}(x)\to f(x)}$ for a.e. ${\displaystyle x}$ and ${\displaystyle f_{k}\leq f}$ a.e. for all ${\displaystyle k}$. Then ${\displaystyle f}$ is measurable, the limit ${\displaystyle \displaystyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu }$ exists, and ${\displaystyle \displaystyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu \;=\;\int _{X}f\,d\mu .}$

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 ${\textstyle f=\sup _{k}f_{k}=\lim _{k\to \infty }f_{k}=\liminf _{k\to \infty }f_{k}}$ almost everywhere. The interchange of limits and integrals is then an easy consequence of Fatou's lemma. One has $${\displaystyle \int _{X}f\,d\mu =\int _{X}\liminf _{k}f_{k}\,d\mu \leq \liminf \int _{X}f_{k}\,d\mu }$$ by Fatou's lemma, and then, since ${\displaystyle \int f_{k}\,d\mu \leq \int f_{k+1}\,d\mu \leq \int fd\mu }$ (monotonicity), $${\displaystyle \liminf \int _{X}f_{k}\,d\mu \leq \limsup _{k}\int _{X}f_{k}\,d\mu =\sup _{k}\int _{X}f_{k}\,d\mu \leq \int _{X}f\,d\mu .}$$ Therefore $${\displaystyle \int _{X}f\,d\mu =\liminf _{k\to \infty }\int _{X}f_{k}\,d\mu =\limsup _{k\to \infty }\int _{X}f_{k}\,d\mu =\lim _{k\to \infty }\int _{X}f_{k}\,d\mu =\sup _{k}\int _{X}f_{k}\,d\mu .}$$

• Infinite series
• Dominated convergence theorem