Quantitative L1 Bounds From Uniform Integrability

From Tail Control to an L1 Estimate

Uniform integrability controls the amount of integral carried by large function values. To obtain an explicit bound on $\int |f-g|\,d\mu$, it is useful to separate three effects: the tails where either function is large, an exceptional set where the functions differ substantially, and the remaining region where their difference is bounded by a chosen tolerance. Each contribution can then be estimated using a different piece of information.

Throughout, let $(X,\mathcal{M},\mu)$ be a finite measure space, so $\mu(X)<\infty$. All functions in this tutorial are real-valued, measurable, and in $L^1(\mu)$; inequalities between functions are understood pointwise, and convergence in measure is with respect to $\mu$. No completeness assumption is needed. For a family $\mathcal{F}\subset L^1(\mu)$, define its tail bound at threshold $K>0$ by

U_{\mathcal{F}}(K) = \sup_{h\in\mathcal{F}} \int_{\{|h|>K\}} |h|\,d\mu .

The strict inequality in the tail set is part of this definition. The family is uniformly integrable if $U_{\mathcal{F}}(K)\to0$ as $K\to\infty$. In estimates below, every function in the family is tested at the same threshold $K$, so each tail integral is bounded by this one supremum.

Truncation and the Main Bound

Definition: For $K>0$, the truncation of a real number $s$ to $[-K,K]$ is $T_K(s)=\max(-K,\min(s,K))$. For a measurable real-valued function $h$, define $T_K(h)(x)=T_K(h(x))$. This is measurable and bounded in absolute value by $K$.

Truncation has two useful properties: it changes $h$ only where $|h|>K$, and it does not increase distances between two inputs. In particular, $|h-T_K(h)|=(|h|-K)_+\le |h|\chi_{\{|h|>K\}}$, where $a_+=\max(a,0)$. Also, $|T_K(s)-T_K(t)|\le |s-t|$ for real $s,t$. To verify the latter, suppose $s\le t$. The function $T_K$ is nondecreasing, and on each of the intervals $(-\infty,-K)$, $[-K,K]$, and $(K,\infty)$, its increase between two points is at most the increase of the input. If the points lie in different intervals, split $[s,t]$ at $-K$ and $K$; summing those same inequalities gives $0\le T_K(t)-T_K(s)\le t-s$. Reversing $s,t$ proves the absolute-value inequality.

Theorem: Let $\mathcal{F}\subset L^1(\mu)$ on a finite measure space $(X,\mathcal{M},\mu)$. For $f,g\in\mathcal{F}$, $K>0$, and $\eta>0$, let $A=\{x\in X:|f(x)-g(x)|>\eta\}$. Then $$ \int_X |f-g|\,d\mu \le 2U_{\mathcal{F}}(K)+\eta\,\mu(X)+2K\,\mu(A). $$

Proof. Insert the truncated functions and apply the triangle inequality pointwise: \[ |f-g|\le |f-T_K(f)|+|T_K(f)-T_K(g)|+|T_K(g)-g|. \] The first term is at most $|f|\chi_{\{|f|>K\}}$, whose integral is at most $U_{\mathcal{F}}(K)$ because $f\in\mathcal{F}$. The third term has integral at most $U_{\mathcal{F}}(K)$ for the same reason applied to $g$. For the middle term, the truncation inequality gives $|T_K(f)-T_K(g)|\le |f-g|$. On $A^c$, the definition of $A$ gives $|f-g|\le\eta$, so this middle term is at most $\eta$ there. On $A$, both truncated values lie in $[-K,K]$, so their difference in absolute value is at most $2K$. Integrating over $A^c$ and $A$ therefore bounds the middle integral by $\eta\mu(X)+2K\mu(A)$. Adding the three estimates proves the result. $\square$

The estimate has a direct interpretation. The first term pays for truncating both functions; the second pays for the ordinary error level $\eta$ across the whole space; the final term pays for the exceptional set, using the bound $2K$ on the truncated difference there. In particular, if $\mu(A)$ is small, the exceptional-set contribution is small for fixed $K$.

Worked Example: Checking the Three Contributions

Take $X=[0,1]$ with Lebesgue measure $m$, let $E=[0,1/10]$, and set $f=2\chi_E$, $g=0$. Use $\mathcal{F}=\{f,g\}$, $K=1$, and $\eta=1/2$.

The tail set for $f$ at $K=1$ is $\{|f|>1\}=E$, and its tail integral is $2m(E)=2(1/10)=1/5$. For $g$, the tail integral is zero. Thus $U_{\mathcal{F}}(1)=1/5$. The exceptional set is $A=\{|f-g|>1/2\}=E$, so $m(A)=1/10$. The theorem gives

\int_0^1 |f-g|\,dm \le 2\left(\frac15\right) +\frac12(1) +2(1)\left(\frac1{10}\right) = \frac25+\frac12+\frac15 = \frac{11}{10}.

Direct computation gives $\int_0^1|f-g|\,dm=2m(E)=1/5$, which is indeed at most $11/10$. The theorem is an upper bound, not an assertion that its right side is sharp. A closer check uses the actual truncation errors: $\int|f-T_1(f)|\,dm=\int_E1\,dm=1/10$, while $\int|g-T_1(g)|\,dm=0$. Also $T_1(f)=\chi_E$ and $T_1(g)=0$, so $\int|T_1(f)-T_1(g)|\,dm=1/10$. These three actual contributions sum to $1/10+0+1/10=1/5$, matching the direct integral. The theorem’s more general bounded-region estimate is $\eta m(X)+2K m(A)=1/2+1/5=7/10$; combined with its two tail bounds $2U_{\mathcal{F}}(1)=2/5$, this gives $11/10$ as above.

Choosing the Parameters for a Desired Error

The estimate gives a quantitative way to make an L1 distance small. The order of choices matters: first choose $K$ to control the uniform tails, then choose $\eta$ to control the ordinary-error term, and finally use convergence in measure to control the exceptional set at that fixed $\eta$. The exceptional-set threshold must be held fixed while its measure is made small.

Corollary: Suppose $(X,\mathcal{M},\mu)$ is finite, $\mathcal{F}\subset L^1(\mu)$ is uniformly integrable, and $f_n,f\in\mathcal{F}$ for every $n$. If $f_n\to f$ in measure, then $\int|f_n-f|\,d\mu\to0$.

Proof. If $\mu(X)=0$, all the integrals in the conclusion are zero. Suppose $\mu(X)>0$, and fix $\varepsilon>0$. Uniform integrability gives a threshold $K>0$ such that $2U_{\mathcal{F}}(K)<\varepsilon/3$. Choose $\eta>0$ so that $\eta\mu(X)<\varepsilon/3$. Since $f_n\to f$ in measure, for this fixed $\eta$ there is an $N$ such that \[ \mu(\{|f_n-f|>\eta\})<\frac{\varepsilon}{6K} \quad\text{for all }n\ge N. \] Apply the theorem to $f_n$ and $f$, both members of $\mathcal{F}$, using this same $K$ and $\eta$. For $n\ge N$, \[ \int|f_n-f|\,d\mu < \frac{\varepsilon}{3}+\frac{\varepsilon}{3} +2K\frac{\varepsilon}{6K} =\varepsilon. \] This is the definition of convergence of the integrals to zero. $\square$

Worked Example: A Numerical Error Bound for a Convergent Sequence

On $[0,1]$ with Lebesgue measure, let $f_n=\chi_{[0,1/n]}$ for $n\ge1$, and let $f=0$. Consider $\mathcal{F}=\{f\}\cup\{f_n:n\ge1\}$. Set $K=2$ and $\eta=1/100$.

Every function in this family takes values in $\{0,1\}$, so $\{|h|>2\}$ is empty for each $h\in\mathcal{F}$. Hence $U_{\mathcal{F}}(2)=0$. For $n=100$, the exceptional set is \[ A=\{|f_{100}-f|>1/100\}=[0,1/100], \] since the value on that interval is $1>1/100$, and the value off it is $0\le1/100$. Thus $m(A)=1/100$. The theorem yields \[ \int_0^1|f_{100}-f|\,dm \le 0+\frac1{100}(1)+2(2)\frac1{100} =\frac1{100}+\frac4{100} =\frac1{20}. \] Directly, the integral is $1\cdot m([0,1/100])=1/100$, so the bound holds. More generally, for every $n$, $\int|f_n-f|\,dm=1/n$; and for any fixed $\eta>0$, once $1>\eta$, the set $\{|f_n-f|>\eta\}$ has measure $1/n\to0$. This verifies convergence in measure, while the exact integrals confirm convergence in $L^1$ in this example.

Why a Tail Hypothesis Cannot Be Dropped

Worked Example: Spikes with Small Exceptional Sets but No Small L1 Error

On $\mathbb{R}$ with Lebesgue measure $m$, define $f_n=n\chi_{(0,1/n)}$ and $f=0$. These are real-valued integrable functions. For each $\eta>0$, when $n>\eta$ the set $\{|f_n-f|>\eta\}$ is $(0,1/n)$, of measure $1/n$; hence $f_n\to0$ in measure. But \[ \int_{\mathbb{R}}|f_n-f|\,dm=n\,m((0,1/n))=n(1/n)=1 \] for every $n$, so the L1 distance does not tend to zero.

The missing control is uniform integrability. For any fixed $K>0$, choose an integer $n>K$. Then $\{|f_n|>K\}=(0,1/n)$, and \[ \int_{\{|f_n|>K\}}|f_n|\,dm=n(1/n)=1. \] For every $n$, the corresponding tail integral is either $0$ or $1$, so taking the supremum over the family $\{f_n:n\ge1\}$ gives $U_{\mathcal{F}}(K)=1$. This is true for every $K>0$, and therefore the tail bound does not tend to zero.

Common mistake. A uniform bound on the integrals $\int|f_n|\,d\mu$ is not a substitute for a small uniform tail bound. In the spike example, $\int|f_n|\,dm=1$ for all $n$, yet at every fixed $K$ some member has tail integral $1$. Replacing $U_{\mathcal{F}}(K)$ by a claim that the integrals are bounded would leave the first term in the L1 estimate uncontrolled. The theorem requires the tail integrals themselves to become uniformly small as the threshold increases.

Check Your Understanding

In each question, keep the threshold $K$ fixed when taking the supremum or applying convergence in measure.

For $h=3\chi_E$, where $\mu(E)=1/12$, compute $\int_{\{|h|>2\}}|h|\,d\mu$.
On a finite measure space, explain why $\mu(\{|f-g|>\eta\})$ controls the bounded-truncation contribution by $\eta\mu(X)+2K\mu(\{|f-g|>\eta\})$.
If $U_{\mathcal{F}}(K)<\varepsilon/8$, what bound does this give for the two tail terms in the theorem?
For $f_n=n\chi_{(0,1/n)}$, compute the L1 distance to zero and the tail integral at a fixed $K>0$ for an integer $n>K$.
Why must the corollary assume that the limit function $f$ belongs to the uniformly integrable family being used in its proof?

Quantitative L1 Bounds From Uniform Integrability

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