Why Stability Matters
Uniform integrability is a property of a whole family, not just of each function separately. When estimating a sequence or a collection of functions, it is useful to know which routine operations preserve that property. A common integrable bound controls every member at once; finite sums and uniformly bounded multipliers also preserve the smallness of large-value tails. The uniformity in these statements is essential: bounds that depend on the individual function need not give a uniform conclusion.
Throughout the main results, \((X,\mathcal{M},\mu)\) is a finite measure space, meaning \(\mu(X)<\infty\). Functions are measurable and may be real-valued or complex-valued; absolute values are used in either case. Integrable means \(\int_X |f|\,d\mu<\infty\). All pointwise bounds between functions may hold \(\mu\)-a.e.; changing values on null sets does not affect the integrals below.
Domination by an Integrable Envelope
An integrable envelope is a single nonnegative integrable function that bounds the absolute values of every member of a family. This condition directly places every large-value tail inside a tail of the envelope.
Proof. Since \(|f|\le g\) a.e., monotonicity of the integral gives \(\int |f|\,d\mu\le\int g\,d\mu<\infty\). For each \(K>0\), the inclusion \(\{|f|>K\}\subseteq\{g>K\}\) holds up to a null set. Therefore $$ \int_{\{|f|>K\}}|f|\,d\mu\le\int_{\{g>K\}}g\,d\mu. $$ The right side does not depend on \(f\). As \(K\to\infty\), \(g\chi_{\{g>K\}}\to0\) a.e., since an integrable nonnegative function is finite a.e.; also \(|g\chi_{\{g>K\}}|\le g\), with \(g\) integrable. The Dominated Convergence Theorem therefore gives \(\int_{\{g>K\}}g\,d\mu\to0\). Taking the supremum over \(f\) proves the asserted uniform tail limit. \(\square\)
Worked Example: A Singular Integrable Envelope
On \((0,1)\) with Lebesgue measure \(m\), set \(g(x)=2x^{-1/2}\), and define \(f_n(x)=\min\{g(x),n\}\) for \(n\ge1\). Verify uniform integrability by computing the envelope tail exactly.
First, \(g\) is integrable and $$ \int_0^1g(x)\,dm(x)=\int_0^1 2x^{-1/2}\,dx=4\sqrt{x}\Big|_0^1=4. $$ For \(K\ge2\), the inequality \(g(x)>K\) is equivalent to \(2/\sqrt{x}>K\), or \(0<x<4/K^2\). Hence $$ \int_{\{g>K\}}g\,dm=\int_0^{4/K^2}2x^{-1/2}\,dx =4\sqrt{x}\Big|_0^{4/K^2}=\frac{8}{K}. $$ Because \(0\le f_n\le g\), the domination theorem applies, and more explicitly, $$ \int_{\{f_n>K\}}f_n\,dm\le\int_{\{g>K\}}g\,dm=\frac{8}{K} \qquad(K\ge2) $$ for every \(n\). The right side tends to zero independently of \(n\), proving uniform integrability. Check: at \(K=2\), \(\{g>2\}=(0,1)\) and the tail integral is \(4\), which agrees with \(8/K=4\). This also checks the factor in the antiderivative.
From Tail Control to Small-Set Control
To handle finite sums, we need a consequence of uniform integrability. On a finite measure space, uniform tail control implies that the integrals over small measurable sets are small uniformly across the family. We prove the implication, since it is the step that controls cross-contributions when a sum is large.
Proof. By uniform integrability, choose \(K\ge1\) such that \(\sup_{f\in\mathcal{F}}\int_{\{|f|>K\}}|f|\,d\mu<\varepsilon/2\). Set \(\delta=\varepsilon/(2K)\). For each \(f\in\mathcal{F}\), split \(E\) into the parts where \(|f|\le K\) and \(|f|>K\). Then $$ \int_E|f|\,d\mu \le K\mu(E)+\int_{\{|f|>K\}}|f|\,d\mu <K\delta+\frac{\varepsilon}{2}=\varepsilon. $$ The choices of \(K\) and \(\delta\) are independent of \(f\), so taking the supremum proves the claim. \(\square\)
This lemma also gives a uniform \(L^1\) bound. Indeed, using a threshold \(K\) with tail integrals at most \(1\), for every \(f\in\mathcal{F}\) we have \(\int |f|\,d\mu\le K\mu(X)+1\). The finiteness of \(\mu(X)\) is used here; the small-set lemma and the finite-sum result below are stated in that setting.
Finite Sums Preserve Uniform Integrability
Proof. For \(K>0\) and a chosen tuple \(f_i\in\mathcal{F}_i\), let \(A=\{|f_1+\cdots+f_r|>K\}\). The triangle inequality implies that \(A\subseteq\bigcup_{j=1}^r\{|f_j|>K/r\}\) up to a null set: if all \(r\) absolute values were at most \(K/r\), their sum would be at most \(K\). Consequently, $$ \mu(A)\le\sum_{j=1}^r\mu(\{|f_j|>K/r\}) \le\frac{r}{K}\sum_{j=1}^r\int_{\{|f_j|>K/r\}}|f_j|\,d\mu. $$ For each \(j\), uniform integrability makes the corresponding tail supremum tend to zero as \(K\to\infty\); in particular those suprema are bounded for all sufficiently large \(K\). The displayed estimate thus gives \(\sup\mu(A)\to0\), where the supremum is over all choices of the tuple. Apply the small-set lemma to each \(\mathcal{F}_i\): given \(\varepsilon>0\), there is a \(\delta_i>0\) such that \(\mu(E)<\delta_i\) implies \(\sup_{f_i\in\mathcal{F}_i}\int_E|f_i|\,d\mu<\varepsilon/r\). Since \(\sup\mu(A)\to0\), for all sufficiently large \(K\), \(\mu(A)<\min_i\delta_i\) for every tuple. It follows that $$ \int_A|f_1+\cdots+f_r|\,d\mu \le\sum_{i=1}^r\int_A|f_i|\,d\mu<\varepsilon. $$ The left side is exactly the tail integral of the sum at threshold \(K\), and the bound is uniform over the tuple. This proves uniform integrability of \(\mathcal{S}\). \(\square\)
Worked Example: Adding Two Spike Families
On \((0,1)\) with Lebesgue measure, let \(f_n=n\chi_{(0,1/n^2)}\) and consider the two families \(\{f_n:n\ge1\}\) and \(\{2f_n:n\ge1\}\). Verify directly that their termwise sums form a uniformly integrable family.
The integral of \(f_n\) is \(n\cdot(1/n^2)=1/n\). If \(n\le K\), then \(\{|f_n|>K\}\) is empty. If \(n>K\), the tail is the whole interval \((0,1/n^2)\), and its integral is \(1/n\). Therefore the tail supremum for the first family is at most \(1/K\) for \(K>0\). For the second family, \(2f_n=2n\chi_{(0,1/n^2)}\) has tail integral zero when \(2n\le K\) and \(2/n<4/K\) when \(2n>K\), so its tail supremum is at most \(4/K\); both families are uniformly integrable.
The sum is \(f_n+2f_n=3n\chi_{(0,1/n^2)}\). Its tail integral at level \(K\) is zero if \(3n\le K\), and is \(3/n\) if \(3n>K\). In the latter case \(3/n<9/K\), so the supremum of the sum family's tail integrals is at most \(9/K\), which tends to zero. Check: for \(n=2\), the sum has height \(6\) on a set of measure \(1/4\), hence integral \(6/4=3/2\), also given by \(3/n=3/2\). Its integral is the sum of the separate integrals \(1/2+1=3/2\).
Multiplication by Uniformly Bounded Functions
Proof. If \(C=0\), each product is zero a.e. and the assertion follows. Suppose \(C>0\). For any \(h\in\mathcal{H}\), \(f\in\mathcal{F}\), and \(K>0\), the set \(\{|hf|>K\}\) is contained, up to a null set, in \(\{|f|>K/C\}\), since \(|h|\le C\). Also \(|hf|\le C|f|\). Thus $$ \int_{\{|hf|>K\}}|hf|\,d\mu \le C\int_{\{|f|>K/C\}}|f|\,d\mu. $$ Taking the supremum over \(f\) and \(h\), the right side tends to zero as \(K\to\infty\), by uniform integrability of \(\mathcal{F}\). The products are integrable because \(|hf|\le C|f|\). \(\square\)
Worked Example: Multiplying Spikes by Bounded Oscillations
Use the spike family \(f_n=n\chi_{(0,1/n^2)}\) on \((0,1)\), and set \(h_n(x)=2(-1)^n\). Compute a uniform tail bound for \(h_nf_n\).
Here \(|h_n|=2\) everywhere, so the multiplier bound is \(C=2\). The product has absolute value \(2n\) on \((0,1/n^2)\) and zero elsewhere. If \(2n\le K\), its tail integral is zero; if \(2n>K\), its tail integral is $$ 2n\cdot\frac{1}{n^2}=\frac{2}{n}<\frac{4}{K}. $$ Thus the supremum of the tail integrals is at most \(4/K\), proving uniform integrability. Check: for \(n=3\) and \(K=4\), the tail integral is \(6\cdot(1/9)=2/3\), which is below \(4/K=1\). Directly, the product's \(L^1\) norm is \(2/3\), the same value.
Common Pitfalls and the Role of Uniform Bounds
A common calculation error in an envelope estimate is to lose a factor when evaluating the tail integral. For \(g(x)=2x^{-1/2}\) on \((0,1)\) and \(K\ge2\), the correct calculation is $$ \int_{\{g>K\}}g\,dm=\int_0^{4/K^2}2x^{-1/2}\,dx =4\sqrt{x}\Big|_0^{4/K^2}=\frac{8}{K}, $$ not \(4/K\). At \(K=2\), the region is all of \((0,1)\), whose integral is \(4\); the incorrect value \(2\) fails this check. The corrected bound still tends to zero and is sufficient for uniform integrability.
Another mistake is to let the multiplier bound depend on the function. For example, on \((0,1)\), the constant functions \(f_n=1\) form a uniformly integrable family: their tail integrals are zero for \(K\ge1\). But multiplying \(f_n\) by the constant multiplier \(h_n=n\) gives the family of constant functions \(n\). For any fixed \(K\), every \(n>K\) has tail integral \(\int_0^1 n\,dm=n\); the supremum is infinite. Each multiplier is bounded individually, but there is no single finite bound for all \(h_n\). The bounded-multiplier theorem requires precisely that common bound.
For finite sums, it is also insufficient merely to observe that each summand is integrable. The proof needs the tail estimates uniformly over each family and the finite-measure small-set control to handle a summand that is not itself large on the set where the entire sum is large. These are the mechanisms behind the stability result, not just formal consequences of finite additivity.
Check Your Understanding
For each question, identify the space and check that the needed uniformity and integrability hypotheses hold.
- If \(|f|\le g\) a.e. for every member of a family and \(g\ge0\) is integrable, explain why \(\{|f|>K\}\subseteq\{g>K\}\) a.e.
- On a finite measure space, use the tail definition to show that a uniformly integrable family has uniformly bounded \(L^1\) norms.
- For two uniformly integrable families, explain why the set where the absolute value of a sum exceeds \(K\) is contained in the union of the sets where one summand exceeds \(K/2\).
- For \(f_n=n\chi_{(0,1/n^2)}\), compute the tail integral at threshold \(K\), separating the cases \(n\le K\) and \(n>K\).
- Give an example showing why individual boundedness of multipliers does not replace one common uniform bound.