Common Errors with Monotone Convergence and Fatou

Start by Checking What the Theorem Requires

Monotone Convergence and Fatou's lemma are powerful because they allow conclusions about integrals from information about a sequence of functions. But neither is an unrestricted rule for moving a limit through an integral. Before using either theorem, check that the functions are measurable and nonnegative; for Monotone Convergence, also check that they increase. Missing one of these assumptions may mean the integral is not defined, the conclusion fails, or the conclusion is an inequality rather than an equality.

Here and below, $ (X,\mathcal{M},\mu) $ is an arbitrary measure space; it need not be finite or sigma-finite. The functions in the theorem statements are extended real-valued and measurable. The following are standard results established in the earlier material on these theorems.

Standard results (Monotone Convergence and Fatou's lemma): If $f_n:X\to[0,\infty]$ are measurable and $f_n\le f_{n+1}$ everywhere for every $n$, then $f=\lim_n f_n$ is measurable and $\int f\,d\mu=\lim_n\int f_n\,d\mu$, with $+\infty$ allowed. If $f_n:X\to[0,\infty]$ are measurable, without a monotonicity assumption, then $\int\liminf_n f_n\,d\mu\le\liminf_n\int f_n\,d\mu$. Both statements also hold when the hypotheses concerning order or pointwise limits hold $\mu$-a.e.

These hypotheses are not interchangeable. Measurability makes the integrals meaningful; nonnegativity is part of both statements in these forms; and only Monotone Convergence requires an increasing sequence. Fatou gives an inequality, not generally an equality.

Error 1: Treating Nonmeasurable Functions as Integrable

A function must be measurable with respect to the stated sigma-algebra for its Lebesgue integral to be defined. For example, let $V\subset[0,1]$ be a non-Lebesgue-measurable set, such as a set obtained by the standard Vitali construction. On $([0,1],\mathcal{L},m)$, the function $\chi_V$ is not measurable. Consequently, neither $\int\chi_V\,dm$ nor the integrals of a constant sequence $f_n=\chi_V$ are Lebesgue integrals.

This is not an example in which a theorem's conclusion has been computed and shown false: the expressions to which the theorem would be applied are not defined as Lebesgue integrals. Replacing $m$ by Lebesgue outer measure $m^*$ does not repair the argument, since outer measure is not the measure used in the Lebesgue integral.

There is a useful check in the other direction: a pointwise limit of measurable functions is measurable. Thus, if a proposed example has measurable $f_n$ but a nonmeasurable pointwise limit, some part of the claim must be wrong.

Proposition: On a measurable space $(X,\mathcal{M})$, if real-valued measurable functions $f_n$ converge pointwise everywhere to a real-valued function $f$, then $f$ is measurable.

Proof. Fix $a\in\mathbb{R}$. For every $x\in X$, convergence $f_n(x)\to f(x)$ implies $f(x)>a$ if and only if there is a rational number $q>a$ and an integer $N\ge1$ such that $f_n(x)>q$ for every $n\ge N$. In the forward direction, choose a rational $q$ strictly between $a$ and $f(x)$; convergence then gives the eventual strict inequality. In the reverse direction, eventual $f_n(x)>q$ implies $f(x)\ge q>a$. Therefore $$ \{x:f(x)>a\} =\bigcup_{\substack{q\in\mathbb{Q}\\q>a}}\ \bigcup_{N=1}^{\infty}\ \bigcap_{n=N}^{\infty}\{x:f_n(x)>q\}. $$ Each set on the right is measurable, and the unions are countable because $\mathbb{Q}$ is countable. Hence $\{f>a\}\in\mathcal{M}$ for every real $a$, which is a measurability criterion for real-valued functions. Thus $f$ is measurable. $\square$

The proposition concerns pointwise convergence everywhere. For convergence only $\mu$-a.e. on a possibly incomplete measure space, the limit as originally specified on the exceptional set need not be measurable. One may instead use a measurable version when the setting permits it; do not silently treat an a.e. limit as an everywhere-defined measurable function without checking.

Error 2: Assuming Pointwise Convergence Is Enough for Equality

Monotone Convergence requires an increasing sequence. If a nonnegative sequence is not increasing, pointwise convergence alone does not justify equality between the integral of the limit and the limit of the integrals.

Worked Example: Pointwise Convergence with No Convergence of Integrals

On $(\mathbb{R},\mathcal{L},m)$, define $f_n(x)=n\chi_{(0,1/n)}(x)$. Each $f_n$ is nonnegative and measurable. At $x=0$, every $f_n(x)=0$. At any fixed $x>0$, once $n\ge 1/x$, we have $x\notin(0,1/n)$, so $f_n(x)=0$; for $x<0$, every value is zero. Thus $f_n(x)\to0$ everywhere and $\int\lim_n f_n\,dm=0$.

The interval $(0,1/n)$ has measure $1/n$, so direct integration gives $$ \int_{\mathbb{R}}f_n\,dm=n\,m((0,1/n))=n\cdot\frac1n=1. $$ It follows that $\lim_n\int f_n\,dm=1$, not $0$. To check that monotonicity really is missing, at $x=3/4$ we have $f_1(3/4)=1$, since $3/4\in(0,1)$, while $f_2(3/4)=0$, since $3/4\notin(0,1/2)$. Therefore $f_1\not\le f_2$. The calculation shows exactly why the equality conclusion of Monotone Convergence cannot be used here.

Check: the pointwise limit has integral $0$, whereas each integral is $1$. These values follow directly from the limit and interval measure computed above.

The same sequence is still within Fatou's hypotheses: its functions are measurable and nonnegative. Fatou gives $0\le1$, which is true. A common error is to treat Fatou's inequality as an equality or as an alternative statement of Monotone Convergence. It is neither.

Worked Example: Fatou's Inequality Can Be Strict

On $(\mathbb{R},\mathcal{L},m)$, set $g_n=\chi_{[n,n+1]}$. These functions are measurable and nonnegative. For each fixed $x\in\mathbb{R}$, at most two of the intervals $[n,n+1]$, with $n\ge1$, contain $x$. Therefore $g_n(x)=0$ for all sufficiently large $n$, so $\liminf_n g_n(x)=0$ everywhere and $\int\liminf_n g_n\,dm=0$.

Each interval $[n,n+1]$ has Lebesgue measure $1$, so $\int g_n\,dm=1$ for every $n$ and $\liminf_n\int g_n\,dm=1$. Thus Fatou reads $0\le1$, and the inequality is strict. Check: the limit inferior of the functions is zero at every point, while the limit inferior of their integrals is $1$; these are the two quantities Fatou compares.

Error 3: Applying Fatou to Signed Functions

In its standard form above, Fatou's lemma requires nonnegative functions. Measurability alone does not suffice. The next example has measurable, integrable functions, but they take negative values; omitting nonnegativity makes the claimed inequality false.

Worked Example: A Signed Sequence Violating Fatou's Inequality

On $(\mathbb{R},\mathcal{L},m)$, define $h_n=-\chi_{[n,n+1]}$. Each $h_n$ is measurable and integrable. For any fixed $x$, only finitely many of the intervals $[n,n+1]$ contain $x$, so $h_n(x)\to0$ everywhere. Consequently, $\int\liminf_n h_n\,dm=0$.

For every $n$, the integral is $$ \int_{\mathbb{R}}h_n\,dm=-m([n,n+1])=-1. $$ Hence $\liminf_n\int h_n\,dm=-1$. The Fatou inequality would assert $0\le-1$, which is false. The functions are measurable, and the integrals are finite; the missing condition is nonnegativity. Check: the interval length is $1$, so changing the sign of its indicator gives integral $-1$, not $1$.

This example is distinct from the nonmeasurability problem: here all the displayed integrals exist, and the proposed inequality itself fails. There are extensions of Fatou's lemma under other hypotheses, but they are not the nonnegative version stated above. Do not use a signed version without stating and checking its additional assumptions.

A Reliable Audit Before Using Either Theorem

For a proposed limit-integral argument on $(X,\mathcal{M},\mu)$, write down the sequence's properties before evaluating anything. Check that each $f_n$ is $\mathcal{M}$-measurable, identify its range and sign, and state whether convergence is pointwise everywhere or $\mu$-a.e. Then choose the conclusion that the hypotheses actually provide:

Monotone Convergence: measurable, nonnegative functions increasing pointwise (or $\mu$-a.e.); conclusion is equality, including possible infinite values.
Fatou: measurable, nonnegative functions; no monotonicity is required, and the conclusion is an inequality.
Neither conclusion follows automatically: the functions are not measurable, or a signed sequence is used without a suitable additional theorem.

Finally, keep the distinction between a theorem not applying and a theorem producing a false conclusion. For the Vitali-set example the integral is undefined, so there is no valid integral calculation to compare. For the moving-interval signed example the integrals are defined, and the claimed Fatou inequality fails. Naming which situation occurs is part of diagnosing the error.

Check Your Understanding

For each question, identify the measure space and check every relevant hypothesis before drawing a conclusion.

On $(\mathbb{R},\mathcal{L},m)$, let $u_n=n\chi_{(0,1/n)}$. Which hypothesis of Monotone Convergence fails, and what are $\int\lim_n u_n\,dm$ and $\lim_n\int u_n\,dm$?
Why is a constant sequence $v_n=\chi_V$, for a non-Lebesgue-measurable $V\subset[0,1]$, not a counterexample showing an incorrect numerical conclusion of Fatou's lemma?
On $(\mathbb{R},\mathcal{L},m)$, let $w_n=-\chi_{[n,n+1]}$. Compute $\int\liminf_n w_n\,dm$ and $\liminf_n\int w_n\,dm$. Which Fatou hypothesis fails?
For $g_n=\chi_{[n,n+1]}$ on $(\mathbb{R},\mathcal{L},m)$, compute both sides of Fatou's inequality and explain why equality is not required.
Can a pointwise limit everywhere of real-valued measurable functions on $(X,\mathcal{M})$ be nonmeasurable? State the criterion that supports your answer.