The Proof Strategy
The Monotone Convergence Theorem was stated in the previous tutorial. Its conclusion is simple to express: when nonnegative measurable functions increase pointwise, their integrals converge to the integral of their pointwise limit. The proof must connect pointwise information at each \(x\) to the integral, which is defined through measurable simple functions below the integrand.
One inequality follows directly from the Monotonicity of the Lebesgue Integral theorem. For the other, fix any nonnegative simple function below the limit. We will show that a fixed fraction of this simple function is eventually supported, level by level, where the approximating functions are large enough. The key is to use a fraction strictly less than one: it creates room for the approximating functions to reach the required threshold, even when they do not reach the full simple-function value at any particular finite stage.
Since \(f_n\leq f\), the Monotonicity of the Lebesgue Integral theorem gives \(\int_X f_n\,d\mu\leq\int_X f\,d\mu\) for every \(n\). The integrals on the left form a nondecreasing sequence, again by monotonicity, so their limit \(L\) exists in \([0,\infty]\). Taking limits in the inequality gives $$ L\leq\int_X f\,d\mu. $$ It remains to prove the reverse inequality.
A Simple Function on Increasing Sets
We first record how integrals behave when a fixed simple function is restricted to a sequence of increasing sets. This is the step that lets pointwise threshold information become an integral estimate.
Proof. Write the positive-level representation of \(h\) as \(h=\sum_{j=1}^{r}a_j\mathbf{1}_{A_j}\), where the \(a_j\) are distinct positive values and the measurable sets \(A_j=\{x:h(x)=a_j\}\) are pairwise disjoint. Since the union of the \(E_n\) contains \(\{h>0\}\), it contains every \(A_j\). Thus \(A_j\cap E_n\) increases to \(A_j\). By Continuity from Below, $$ \mu(A_j\cap E_n)\longrightarrow\mu(A_j) $$ for each \(j\). The simple-function integral formula gives $$ \int_X h\mathbf{1}_{E_n}\,d\mu =\sum_{j=1}^{r}a_j\mu(A_j\cap E_n), \qquad \int_X h\,d\mu=\sum_{j=1}^{r}a_j\mu(A_j). $$ There are only finitely many terms, so taking limits term by term gives the claimed equality. This remains true if one of the measures is infinite: its corresponding term tends to \(+\infty\), and so does the finite sum. \(\square\)
The lemma does not require the sets \(E_n\) to cover all of \(X\). They need only cover the region where \(h\) is positive. This matters because the simple function we use below may vanish on much of the space.
Proof of the Monotone Convergence Theorem
Proof. Let \(L=\lim_{n\to\infty}\int_X f_n\,d\mu\). We have already shown that \(L\leq\int_X f\,d\mu\). For the reverse inequality, take any nonnegative measurable simple function \(h\) satisfying \(h\leq f\), and fix a real number \(\alpha\) with \(0<\alpha<1\). Define $$ E_n=\{x\in X:h(x)>0\text{ and }f_n(x)\geq\alpha h(x)\}. $$ These sets are measurable. For example, since \(h\) has only finitely many values, \(E_n\) is a finite union of intersections of level sets of \(h\) with measurable threshold sets of \(f_n\). They are increasing because \(f_n\leq f_{n+1}\).
Every point where \(h\) is positive eventually belongs to \(E_n\). To verify this, fix such a point \(x\). Then \(h(x)>0\), and \(f(x)\geq h(x)>\alpha h(x)\). Since \(f_n(x)\) increases to \(f(x)\), some \(n\) satisfies \(f_n(x)\geq\alpha h(x)\). This also holds if \(f(x)=+\infty\). Therefore \(\bigcup_{n=1}^{\infty}E_n\) contains \(\{h>0\}\).
On \(E_n\), the defining inequality gives \(\alpha h\leq f_n\); off \(E_n\), the function \(\alpha h\mathbf{1}_{E_n}\) is zero and \(f_n\geq0\). Thus \(\alpha h\mathbf{1}_{E_n}\leq f_n\) everywhere. Monotonicity of the integral and the Simple Functions on Increasing Sets lemma now give $$ \int_X f_n\,d\mu \geq \alpha\int_X h\mathbf{1}_{E_n}\,d\mu, \qquad \lim_{n\to\infty}\int_X h\mathbf{1}_{E_n}\,d\mu =\int_X h\,d\mu. $$ Taking limits in the first inequality yields \(L\geq\alpha\int_X h\,d\mu\).
If \(\int_X h\,d\mu=+\infty\), this inequality implies \(L=+\infty\). If \(\int_X h\,d\mu\) is finite, the inequality holds for every \(0<\alpha<1\); letting \(\alpha\) increase to \(1\) gives \(L\geq\int_X h\,d\mu\). Thus in either case \(L\geq\int_X h\,d\mu\) for every nonnegative measurable simple \(h\leq f\). By the definition of the nonnegative integral as the supremum of these simple-function integrals, $$ L\geq\sup_{\substack{h\text{ nonnegative measurable simple}\\h\leq f}}\int_X h\,d\mu =\int_X f\,d\mu. $$ Together with \(L\leq\int_X f\,d\mu\), this proves the theorem. \(\square\)
Worked Examples of the Proof’s Ingredients
Worked Example: Dyadic Simple Functions Approaching the Identity
On \([0,1]\) with Lebesgue measure, define \(s_n(1)=1\), and, for \(0\leq x<1\), set $$ s_n(x)=\frac{\lfloor 2^n x\rfloor}{2^n}. $$ Each \(s_n\) is a nonnegative measurable simple function. The dyadic intervals of length \(2^{-n}\) partition \([0,1)\), and \(s_n\) takes the value \(j/2^n\) on \([j/2^n,(j+1)/2^n)\), for \(j=0,\ldots,2^n-1\).
For \(x<1\), write \(j=\lfloor 2^n x\rfloor\). Then \(j\leq 2^n x<j+1\), so \(2j\leq2^{n+1}x<2j+2\). It follows that \(\lfloor2^{n+1}x\rfloor\) is either \(2j\) or \(2j+1\), and hence \(s_{n+1}(x)\geq j/2^n=s_n(x)\). At \(x=1\), both values are \(1\). Also \(0\leq x-s_n(x)<2^{-n}\) for \(x<1\), so \(s_n(x)\to x\); at \(x=1\), the sequence is constantly \(1\).
Each interval in the partition has measure \(2^{-n}\). The simple-function integral formula therefore gives $$ \int_{[0,1]}s_n\,d\lambda =\sum_{j=0}^{2^n-1}\frac{j}{2^n}\,2^{-n} =\frac{1}{2^{2n}}\cdot\frac{(2^n-1)2^n}{2} =\frac{1-2^{-n}}{2}. $$ The limit of these integrals is \(1/2\). The Monotone Convergence Theorem gives \(\int_{[0,1]}x\,d\lambda=1/2\). Here the proof’s simple-function viewpoint is explicit: each stage supplies a lower estimate whose integral approaches the integral of the limit.
Worked Example: Threshold Sets on a Finite Atomic Space
Let \(X=\{u,v,w\}\), let every subset be measurable, and assign masses \(\mu(\{u\})=2\), \(\mu(\{v\})=3\), and \(\mu(\{w\})=5\). For \(n\geq1\), define $$ f_n(u)=\frac{n}{n+1},\qquad f_n(v)=\frac{2n}{n+1},\qquad f_n(w)=0. $$ The sequence increases at each point and converges to \(f(u)=1\), \(f(v)=2\), and \(f(w)=0\). Its integrals are $$ \int_X f_n\,d\mu =2\frac{n}{n+1}+3\frac{2n}{n+1} =\frac{8n}{n+1}\longrightarrow 8. $$ The limit function has integral \(2\cdot1+3\cdot2+5\cdot0=8\).
The threshold-set argument can be seen directly with the simple function \(h=f\) and \(\alpha=3/4\). At \(u\), the threshold is \(3/4\); the inequality \(n/(n+1)\geq3/4\) is equivalent to \(4n\geq3n+3\), or \(n\geq3\). At \(v\), the threshold is \(3/2\); \(2n/(n+1)\geq3/2\) is equivalent to \(4n\geq3n+3\), again true exactly when \(n\geq3\). Thus \(E_n\) contains neither \(u\) nor \(v\) for \(n=1,2\), and contains both for every \(n\geq3\). The sets eventually cover the positive support of \(h\), just as the proof requires.
Worked Example: An Infinite Limiting Integral
Take \(X=\mathbb{N}\) with counting measure and define \(f_n(k)=k\) if \(1\leq k\leq n\), and \(f_n(k)=0\) if \(k>n\). Each \(f_n\) is a nonnegative measurable simple function, and \(f_n(k)\leq f_{n+1}(k)\) for every \(k\). For fixed \(k\), the value is eventually \(k\), so the pointwise limit is \(f(k)=k\).
The integral of \(f_n\) is its finite sum of values: $$ \int_{\mathbb{N}}f_n\,d\mu =\sum_{k=1}^{n}k =\frac{n(n+1)}{2}\longrightarrow+\infty. $$ The limiting integral is also infinite. Indeed, for each \(m\), the simple function equal to \(k\) on \(1\leq k\leq m\) and zero elsewhere lies below \(f\), and its integral is \(m(m+1)/2\). These lower bounds are unbounded as \(m\to\infty\), so the defining supremum for \(\int_{\mathbb{N}}f\,d\mu\) is \(+\infty\). The theorem therefore gives equality even when both sides are infinite.
Why the Strict Fraction Matters
The factor \(\alpha<1\) is essential to this proof. Although \(h(x)\leq f(x)\), it need not be true that \(f_n(x)\geq h(x)\) for any finite \(n\). For example, an increasing sequence of real numbers can approach a limit from below without ever attaining it. But if \(h(x)>0\), then \(\alpha h(x)<h(x)\leq f(x)\), so the sequence must eventually cross the smaller threshold \(\alpha h(x)\).
There is also no claim that one index works for every point at once. The index at which \(f_n(x)\geq\alpha h(x)\) may depend on \(x\). The increasing sets \(E_n\) handle this pointwise variation: their union covers the positive support of \(h\), and Continuity from Below lets the integral recover the full contribution of \(h\). Requiring a single uniform index would be a stronger condition than pointwise increase provides.
Check Your Understanding
Use the proof’s simple-function and threshold-set steps to check your understanding.
- Which inequality follows immediately from \(f_n\leq f\) and monotonicity of the integral?
- In the Simple Functions on Increasing Sets lemma, why is it enough that \(\bigcup_n E_n\) contain \(\{h>0\}\)?
- Why does \(f(x)\geq h(x)>0\) imply that \(f_n(x)\geq\alpha h(x)\) eventually when \(0<\alpha<1\)?
- Why can the limits in the simple-function lemma be taken term by term, even if a level set has infinite measure?
- In the finite atomic example, from which index onward do the threshold sets with \(\alpha=3/4\) contain both \(u\) and \(v\)?
- How do the finite-stage integrals in the counting-measure example establish that the limiting integral is infinite?