From Fatou’s Inequality to Convergence of Errors
Fatou’s Lemma gives a lower bound for the integral of a pointwise lower limit. One useful application is to compare a sequence of nonnegative functions with its limit when their integrals also converge. Under the right hypotheses, Fatou’s inequality forces the integral of the pointwise minimum to converge to the integral of the limit. That fact controls the error between the functions.
The resulting theorem is often called Scheffé’s Lemma. It says that almost-everywhere convergence, together with convergence of the integrals to the integral of the limit, implies convergence of the absolute errors in integral. The proof illustrates a careful use of Fatou’s Lemma: it does not subtract infinite quantities, and it allows finitely many early functions to have infinite integrals.
Scheffé’s Lemma
Here the functions are finite-valued, but their integrals are initially allowed to be infinite. The convergence of the integrals to a finite number guarantees that \(\int_X f_n\,d\mu\) is finite for all sufficiently large \(n\); it does not guarantee finiteness for every \(n\). The proof will use subtraction only after restricting to those sufficiently large indices.
Proof. Define the pointwise minimum $$ h_n(x)=\min(f_n(x),f(x)). $$ The minimum of two nonnegative measurable functions is measurable, and \(0\leq h_n\leq f\). At every point where \(f_n(x)\to f(x)\), we have \(h_n(x)\to f(x)\). Thus \(h_n\to f\) almost everywhere. Fatou’s Lemma and monotonicity of the integral give $$ \int_X f\,d\mu \leq \liminf_{n\to\infty}\int_X h_n\,d\mu \leq \int_X f\,d\mu. $$ The second inequality follows because \(\int_X h_n\,d\mu\leq\int_X f\,d\mu\) for every \(n\). Consequently, $$ \lim_{n\to\infty}\int_X h_n\,d\mu=\int_X f\,d\mu. $$ In particular, every \(\int_X h_n\,d\mu\) is finite.
Write \(L=\int_X f\,d\mu\), which is finite. Since \(\int_X f_n\,d\mu\to L\), there is an integer \(N\) such that \(\int_X f_n\,d\mu<\infty\) for every \(n\geq N\). Indeed, for all sufficiently large \(n\), these integrals are within \(1\) of \(L\), and hence are finite. We now fix \(n\geq N\). The pointwise identities $$ f_n=h_n+(f_n-h_n),\qquad f=h_n+(f-h_n) $$ involve only nonnegative measurable functions. Additivity of the nonnegative integral, and finiteness of the integrals on the left and of \(\int_X h_n\,d\mu\), yield $$ \int_X(f_n-h_n)\,d\mu = \int_X f_n\,d\mu-\int_X h_n\,d\mu $$ and $$ \int_X(f-h_n)\,d\mu = \int_X f\,d\mu-\int_X h_n\,d\mu. $$ These subtractions are between finite numbers. Since $$ |f_n-f|=(f_n-h_n)+(f-h_n), $$ another application of additivity gives $$ \int_X|f_n-f|\,d\mu = \int_X f_n\,d\mu+\int_X f\,d\mu -2\int_X h_n\,d\mu. $$ As \(n\to\infty\), each of the first two terms on the right tends to \(L\), and the last integral tends to \(L\). Therefore the right-hand side tends to \(L+L-2L=0\). The finitely many indices before \(N\) do not affect this limit. This proves the theorem.
Worked Examples
Worked Example: Powers on the Unit Interval
Take \(X=[0,1]\) with Lebesgue measure, let \(f_n(x)=x^n\), and define \(f(x)=0\) for \(0\leq x<1\) and \(f(1)=1\). For \(x<1\), \(x^n\to0\), while \(f_n(1)=1\) for every \(n\). Thus \(f_n\to f\) everywhere. The function \(f\) vanishes except at one point, so its integral is zero. Direct integration gives $$ \int_0^1 f_n(x)\,dx = \left[\frac{x^{n+1}}{n+1}\right]_0^1 = \frac{1}{n+1} \longrightarrow 0 = \int_0^1 f(x)\,dx. $$ Scheffé’s Lemma applies. Indeed, except at \(x=1\), \(|f_n(x)-f(x)|=x^n\); at \(x=1\), the difference is zero. A single point has measure zero, so $$ \int_0^1|f_n-f|\,dx = \int_0^1x^n\,dx = \frac{1}{n+1} \longrightarrow0. $$ This example has pointwise convergence even at the endpoint, where the limit has a nonzero value, without changing the integral conclusion.
Worked Example: An Infinite Integral Among the Initial Terms
On \(X=(0,1)\) with Lebesgue measure, let \(f(x)=1\), set \(f_1(x)=1/x\), and set \(f_n(x)=1\) for \(n\geq2\). Every function is finite-valued and measurable, and \(f_n\to f\) everywhere. The first integral is infinite: $$ \int_0^1 f_1(x)\,dx = \int_0^1\frac{1}{x}\,dx = \infty. $$ For every \(n\geq2\), however, $$ \int_0^1 f_n(x)\,dx=1=\int_0^1f(x)\,dx. $$ Thus the integrals converge to the finite integral of \(f\), despite the infinite integral of the first term. For \(n\geq2\), the functions agree pointwise, so $$ \int_0^1|f_n-f|\,dx=0. $$ The example shows why the proof uses finiteness only for all sufficiently large indices rather than claiming that every \(\int_X f_n\,d\mu\) is finite.
Worked Example: A Small Multiplicative Perturbation
On \([0,1]\) with Lebesgue measure, let \(f(x)=1\) and \(f_n(x)=1+1/n\). These are nonnegative measurable functions and \(f_n\to f\) at every point. Their integrals are $$ \int_0^1 f(x)\,dx=1,\qquad \int_0^1 f_n(x)\,dx = \int_0^1\left(1+\frac1n\right)\,dx = 1+\frac1n \longrightarrow1. $$ The absolute error is the constant \(1/n\), so its integral is $$ \int_0^1|f_n-f|\,dx = \int_0^1\frac1n\,dx = \frac1n \longrightarrow0. $$ Here the convergence of the integrals and the convergence of the functions are both transparent. Scheffé’s Lemma guarantees the same conclusion in settings where the absolute error is not so easy to compute directly.
Why Convergence of the Integrals Is Essential
Almost-everywhere convergence by itself does not control the integrals of nonnegative functions. For instance, on \((0,1)\), define \(g_n(x)=n\) for \(0<x<1/n\) and \(g_n(x)=0\) otherwise. For each fixed \(x>0\), eventually \(x\geq1/n\), so \(g_n(x)\to0\). Nevertheless, $$ \int_0^1g_n(x)\,dx = n\cdot\frac1n = 1 $$ for every \(n\). The pointwise limit has integral zero, but the integrals do not converge to that integral. Scheffé’s Lemma does not apply: its integral-convergence hypothesis fails.
The minimum \(h_n=\min(f_n,f)\) is the key comparison in Scheffé’s proof. It lies below both functions and approaches \(f\) almost everywhere. Fatou’s Lemma supplies a lower bound for its limiting integral, while \(h_n\leq f\) supplies the matching upper bound. Once the integrals of the minima are controlled, the two nonnegative parts of the absolute error can be handled using additivity. The finite-integral stage matters: subtracting extended integrals such as \(\infty-\infty\) would be invalid.
Convergence in Measure as a Further Consequence
Proof. By Scheffé’s Lemma, \(\int_X|f_n-f|\,d\mu\to0\). Fix \(\varepsilon>0\) and define the measurable set $$ A_n=\{x\in X:|f_n(x)-f(x)|>\varepsilon\}. $$ On \(A_n\), \(|f_n-f|\geq\varepsilon\), so pointwise on \(X\) we have $$ \varepsilon\mathbf{1}_{A_n}\leq |f_n-f|. $$ Monotonicity of the integral and the integral formula for indicator functions give $$ \varepsilon\,\mu(A_n) = \int_X\varepsilon\mathbf{1}_{A_n}\,d\mu \leq \int_X|f_n-f|\,d\mu. $$ The right-hand side tends to zero. Since \(\varepsilon\) is fixed and positive, \(\mu(A_n)\to0\). Thus \(\mu(\{|f_n-f|>\varepsilon\})\to0\) for every \(\varepsilon>0\); this property is called convergence in measure and is studied in detail later in the course.
This implication adds a useful stability conclusion: under Scheffé’s hypotheses, the error becomes small not only in integral but also outside sets whose measure tends to zero. The converse is not asserted here; convergence in measure alone does not generally give convergence of the integrals of absolute errors.
Check Your Understanding
Use the proof and examples above to answer these questions.
- Why does Fatou’s Lemma give a lower bound for \(\liminf_n\int_X\min(f_n,f)\,d\mu\)?
- Why must the proof allow some initial functions to have infinite integrals?
- At what point in the proof is subtraction used, and why are the quantities being subtracted finite there?
- Why do the functions \(g_n=n\mathbf{1}_{(0,1/n)}\) show that almost-everywhere convergence alone is insufficient?
- How does the bound \(\varepsilon\mu(A_n)\leq\int_X|f_n-f|\,d\mu\) establish convergence in measure?