Why Nonnegativity Matters in Fatou's Lemma

The Lower-Bound Inequality and Its Hypothesis

Fatou’s lemma gives a lower bound for the integral of a pointwise lower limit. Its nonnegativity condition is essential: if functions are allowed to take negative values, a narrow negative spike can have a fixed negative integral even while its support shrinks until the sequence converges pointwise to zero. This tutorial examines that mechanism on a finite interval, where every function in the examples is integrable and every integral is finite.

For the general statement, let $(X,\mathcal{M},\mu)$ be any measure space, not necessarily finite or complete. A function $f:X\to[0,\infty]$ is nonnegative and extended-real-valued. The pointwise lower limit of a sequence is $\liminf_{n\to\infty}f_n=\sup_k\inf_{n\ge k}f_n$. Fatou’s lemma asserts the following.

Theorem (Fatou’s Lemma): If $f_n:X\to[0,\infty]$ are measurable on $(X,\mathcal{M},\mu)$, then $\liminf_n f_n$ is measurable and $$ \int_X \liminf_{n\to\infty} f_n\,d\mu \le \liminf_{n\to\infty}\int_X f_n\,d\mu. $$ Both sides may equal $+\infty$.

Proof. For each $k\ge1$, define $g_k(x)=\inf_{n\ge k}f_n(x)$. A countable infimum of measurable extended-real-valued functions is measurable, so $g_k$ is measurable. Nonnegativity of every $f_n$ gives $g_k\ge0$. Also $g_k\le g_{k+1}$, because the infimum is taken over a smaller tail for $g_{k+1}$, and $\lim_k g_k=\liminf_n f_n$ pointwise. By the Monotone Convergence Theorem, $\int\liminf_n f_n\,d\mu=\lim_k\int g_k\,d\mu$. For every $n\ge k$, $g_k\le f_n$, so monotonicity of the nonnegative integral gives $\int g_k\,d\mu\le\inf_{n\ge k}\int f_n\,d\mu$. Taking $k\to\infty$, the right side tends to $\liminf_n\int f_n\,d\mu$, proving the inequality. $\square$

Here is the point at which the hypothesis enters the proof: it ensures $g_k\ge0$, so the Monotone Convergence Theorem applies to $g_k$. The next examples show that this is not a technicality. The underlying measure space in each example is $([0,1],\mathcal{L}_{[0,1]},m)$, where $\mathcal{L}_{[0,1]}$ is the Lebesgue sigma-algebra restricted to $[0,1]$, and $m$ is Lebesgue measure. This is a finite, complete measure space with $m([0,1])=1$. All functions displayed below are real-valued, measurable, and integrable: each is a bounded simple function for each fixed $n$, with support of finite measure.

A Negative Spike with Fixed Integral

Worked Example: A Negative Spike Converging to Zero

For $n\ge1$, define $f_n(x)=-n\chi_{(0,1/n)}(x)$ on $[0,1]$. Determine its pointwise limit, its integral, and whether the Fatou inequality holds.

At $x=0$, every indicator is zero because the interval is open, so $f_n(0)=0$. Fix $x\in(0,1]$. Once $n\ge1/x$, we have $1/n\le x$, hence $x\notin(0,1/n)$ and $f_n(x)=0$. Thus $f_n(x)\to0$ at every $x\in[0,1]$, and $\liminf_n f_n=0$ everywhere. For each $n$, the interval $(0,1/n)$ has measure $1/n$, so

\int_{[0,1]} f_n\,dm=(-n)m((0,1/n))=(-n)\frac1n=-1.

Consequently, $\int\liminf_n f_n\,dm=0$, whereas $\liminf_n\int f_n\,dm=-1$. The proposed signed inequality would say $0\le-1$, which is false. Check: the integral of the limit is $\int_{[0,1]}0\,dm=0$; independently, the defining simple-function formula gives $(-n)(1/n)+0\cdot(1-1/n)=-1$ for every $n$.

The pointwise limit does not register the spike’s depth: each fixed point is eventually outside its support. The integral, however, multiplies the height by the support’s measure. Here the height is $n$, the width is $1/n$, and their product remains $1$, with a negative sign.

Proposition: On $([0,1],\mathcal{L}_{[0,1]},m)$, let $c>0$, let $r_n\in(0,1]$ satisfy $r_n\to0$, and set $h_n=-(c/r_n)\chi_{(0,r_n)}$. Then $h_n\to0$ pointwise, $\int h_n\,dm=-c$ for every $n$, and $$ \int\liminf_n h_n\,dm=0>-c=\liminf_n\int h_n\,dm. $$

Proof. At $x=0$, $h_n(0)=0$ for all $n$. At any fixed $x\in(0,1]$, convergence $r_n\to0$ gives an $N$ such that $r_n\le x$ for all $n\ge N$. Since the interval is open, $x\notin(0,r_n)$ for those $n$, and $h_n(x)=0$. Thus $h_n(x)\to0$ at every point. Each interval $(0,r_n)$ has measure $r_n$, so $\int h_n\,dm=(-c/r_n)r_n=-c$. The pointwise lower limit is zero, whose integral is zero. Since $c>0$, $0>-c$, proving the strict failure. $\square$

The Same Mechanism with a Nonzero Limit

Worked Example: A Negative Spike Added to a Constant

On the same finite Lebesgue measure space, define $q_n(x)=2-2n\chi_{(0,1/n)}(x)$. Compute the pointwise lower limit and compare the two sides of the signed Fatou inequality.

At $x=0$, $q_n(0)=2$ for every $n$. For fixed $x>0$, once $n\ge1/x$, the indicator is zero, so $q_n(x)=2$. Therefore $q_n\to2$ pointwise, and $\int\liminf_n q_n\,dm=\int_{[0,1]}2\,dm=2$. Using $m((0,1/n))=1/n$ and $m([0,1]\setminus(0,1/n))=1-1/n$, we calculate

\int q_n\,dm =(-2n)\frac1n+2\left(1-\frac1n\right) =-2+2-\frac2n =-\frac2n.

Hence $\liminf_n\int q_n\,dm=0$, and the signed inequality would assert $2\le0$, which fails. Check: for $n=2$, the spike contributes $(-4)(1/2)=-2$, while the rest contributes $2(1/2)=1$, giving $\int q_2\,dm=-1=-2/2$, as in the formula.

This version separates the limiting function from the spike: the sequence converges everywhere to the constant $2$, but the negative spikes subtract an amount approaching $2$ from each integral. In the first example the limit was zero; a zero limit is not needed for the failure.

Why the Nonnegative Comparison Works

Worked Example: A Positive Spike on the Same Intervals

Define $p_n(x)=n\chi_{(0,1/n)}(x)$ on $[0,1]$. Check the pointwise limit and the Fatou inequality.

At $x=0$, $p_n(0)=0$. For every fixed $x>0$, when $n\ge1/x$ the point lies outside $(0,1/n)$, so $p_n(x)=0$ eventually. Thus $p_n\to0$ pointwise and $\int\liminf_n p_n\,dm=0$. For each $n$,

\int p_n\,dm=n\,m((0,1/n))=n\frac1n=1.

Therefore $0=\int\liminf_n p_n\,dm\le1=\liminf_n\int p_n\,dm$, as Fatou’s lemma requires. Check: at $n=4$, the value $4$ occurs on a set of measure $1/4$, giving integral $4(1/4)=1$; the limit function is still zero everywhere.

The positive and negative sequences have the same shrinking supports and the same absolute spike heights. The difference is the direction of the inequality: Fatou gives a lower bound, and a positive spike leaves the limiting integral below the sequence of integrals. A negative spike can push those integrals below the integral of the pointwise lower limit.

The Proof Step That Breaks for Signed Functions

A common incorrect argument is to repeat the tail-infimum proof for a real-valued sequence without checking that its tail infima are nonnegative. For the sequence $f_n=-n\chi_{(0,1/n)}$, one can still define $g_k=\inf_{n\ge k}f_n$, and the tail infima are increasing in $k$. But they can be negative. For example, at any $x\in(0,1/k)$, the $n=k$ term is $f_k(x)=-k$, so $g_k(x)\le-k<0$. The nonnegative Monotone Convergence Theorem used in Fatou’s proof therefore cannot be applied to $(g_k)$.

The invalid calculation is to write “$g_k\uparrow\liminf f_n$, so $\int g_k\,dm\to\int\liminf f_n\,dm$” without verifying $g_k\ge0$. Monotone increase alone does not justify that equality for signed functions. In this example it would lead to the false conclusion $0\le-1$, already contradicted by the direct integral calculation.

Takeaway: Fatou’s lemma is a theorem for nonnegative measurable functions $X\to[0,\infty]$ on a measure space $(X,\mathcal{M},\mu)$. It is not a general lower-bound inequality for arbitrary real-valued functions. On a finite Lebesgue interval, negative spikes of height $c/r_n$ on intervals of width $r_n\to0$ converge pointwise to zero while retaining integral $-c$.

When checking a proposed use of Fatou’s lemma, verify the sign of every function before forming tail infima. Pointwise convergence and integrability of each individual function do not replace the nonnegativity hypothesis: all the counterexample functions above are integrable, and their integrals are finite. The failure comes from the negative mass persisting on sets whose measures shrink to zero.

Check Your Understanding

For each question, state the pointwise limit separately from the sequence of integrals.

On $[0,1]$, let $u_n=-3n\chi_{(0,1/n)}$. Find $\liminf_n u_n$ and $\liminf_n\int u_n\,dm$. Does the signed Fatou inequality hold?
For $v_n=-\sqrt n\,\chi_{(0,1/n)}$, compute the integral and determine whether this sequence violates the lower-bound inequality.
Let $w_n=1-n\chi_{(0,1/n)}$. Find its pointwise limit and the limit of its integrals.
In the proof of Fatou’s lemma, define $g_k=\inf_{n\ge k}f_n$. Which property of $g_k$ follows from nonnegativity, and why is it needed?
For $z_n=4n\chi_{(0,1/n)}$, verify the pointwise limit and check the direction of Fatou’s inequality.