Transforming a Limit in Measure
Convergence in measure controls the size of the set where two functions differ by more than a fixed tolerance. A uniformly continuous function turns sufficiently small input differences into small output differences, with one input tolerance that works everywhere on the real line. This global control is exactly what allows us to transfer convergence in measure through the transformation, even when the functions take large values or the underlying measure space has infinite measure.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; it need not be finite, complete, or sigma-finite. All functions in the main result are real-valued and \(\mathcal{M}\)-measurable, and all statements of convergence in measure use the following definition.
The strict inequality in the error set matters: points where the difference equals the chosen threshold are not counted. No almost-everywhere or uniform convergence is assumed in this definition.
The Uniform-Continuity Theorem
Uniform continuity means that for every \(\varepsilon>0\), there is a \(\delta>0\) such that for all \(s,t\in\mathbb{R}\), \(|s-t|<\delta\) implies \(|\Phi(s)-\Phi(t)|<\varepsilon\). The same \(\delta\) must work for every pair \(s,t\); it cannot depend on where the pair lies on the real line.
Proof. A uniformly continuous real-valued function is continuous, hence Borel measurable. Therefore \(\Phi\circ f_n\) and \(\Phi\circ f\) are measurable. Fix any \(\varepsilon>0\), and choose \(\delta>0\) from uniform continuity for this \(\varepsilon\). If \(|f_n(x)-f(x)|\le\delta/2\), then \(|f_n(x)-f(x)|<\delta\), so \(|\Phi(f_n(x))-\Phi(f(x))|<\varepsilon\). Taking the contrapositive gives the error-set inclusion $$ \{|\Phi(f_n)-\Phi(f)|>\varepsilon\} \subseteq \{|f_n-f|>\delta/2\}. $$ Consequently, $$ 0\le\mu(\{|\Phi(f_n)-\Phi(f)|>\varepsilon\}) \le\mu(\{|f_n-f|>\delta/2\})\longrightarrow 0. $$ The last limit follows from \(f_n\to f\) in measure, applied at the fixed positive threshold \(\delta/2\). Since this holds for every \(\varepsilon>0\), the transformed sequence converges in measure. \(\square\)
This argument needs no bound on \(\mu(X)\): it compares each transformed error set to an original error set whose measure tends to zero. It also explains why a fixed input tolerance is crucial. For each output threshold, uniform continuity supplies one input threshold valid at every point \(x\), no matter how large \(f_n(x)\) or \(f(x)\) may be.
Absolute Value and Truncation
Two common transformations satisfy the theorem, and in both cases there is also a direct quantitative bound. The reverse triangle inequality gives $$ \bigl||s|-|t|\bigr|\le |s-t| \qquad (s,t\in\mathbb{R}), $$ so \(\Phi(t)=|t|\) is uniformly continuous. For \(M>0\), define the truncation \(T_M:\mathbb{R}\to[-M,M]\) by $$ T_M(t)=\max(-M,\min(t,M)). $$ This clamps values below \(-M\) to \(-M\), values above \(M\) to \(M\), and leaves values in \([-M,M]\) unchanged. The map is also uniformly continuous; indeed, it is 1-Lipschitz: $$ |T_M(s)-T_M(t)|\le |s-t|. $$
To verify the last bound, assume \(s\le t\). The map \(T_M\) is nondecreasing. Split the interval \([s,t]\) at either of the points \(-M,M\) that it contains. On each resulting subinterval, \(T_M\) is either constant or the identity, so its increase on that subinterval is at most the subinterval's length. Adding these increases gives \(0\le T_M(t)-T_M(s)\le t-s\). Reversing the roles of \(s,t\) gives the claimed absolute-value inequality. If \(M=0\), the constant map \(T_0=0\) is uniformly continuous as well.
Worked Example: Applying Absolute Value
On \(X=[0,1]\) with Lebesgue measure \(m\) and the Lebesgue sigma-algebra, let \(f_n(x)=x+1/n\) and \(f(x)=x\). These are measurable real-valued functions. Determine the error-set measures for the absolute-value transforms.
Since \(x\ge0\) on \([0,1]\), \(|f_n(x)|=x+1/n\) and \(|f(x)|=x\). Thus the transformed difference is exactly \(1/n\) at every \(x\in[0,1]\). For a fixed \(\varepsilon>0\), $$ \{x:||f_n(x)|-|f(x)||>\varepsilon\} = \begin{cases} [0,1],&1/n>\varepsilon,\\ \varnothing,&1/n\le\varepsilon. \end{cases} $$ The measures are therefore \(1\) in the first case and \(0\) in the second. Once \(n\ge 1/\varepsilon\), the strict error condition fails everywhere, so the measures tend to zero. For example, at \(\varepsilon=1/10\), \(n=5\) gives difference \(1/5>1/10\) and error-set measure \(1\), while \(n=10\) gives difference \(1/10\), which does not exceed the threshold, and error-set measure \(0\).
Check: The original difference is also \(1/n\) everywhere, so its error set has exactly the same measure for every threshold. This agrees with the direct bound \(\bigl||f_n|-|f|\bigr|\le|f_n-f|\), and verifies the strict-threshold calculation.
Worked Example: Truncating a Spike Sequence
On \(X=[0,1]\) with Lebesgue measure, let \(f_n=n\chi_{(0,1/n)}\) and \(f=0\). These functions are measurable and real-valued. First check convergence in measure, then truncate at \(M=2\).
For \(\varepsilon>0\), the original error set is \((0,1/n)\) if \(n>\varepsilon\), because the difference there is \(n>\varepsilon\); otherwise it is empty. Its measure is \(1/n\) when \(n>\varepsilon\), and \(0\) when \(n\le\varepsilon\). For each fixed \(\varepsilon\), all sufficiently large \(n\) satisfy \(n>\varepsilon\), and then \(1/n\to0\). Hence \(f_n\to0\) in measure.
For \(n>2\), truncation gives \(T_2(f_n)=2\chi_{(0,1/n)}\) and \(T_2(f)=0\). At the fixed threshold \(\varepsilon=1\), the transformed error set is \((0,1/n)\), since the transformed difference is \(2>1\) there and \(0\) elsewhere. Its measure is \(1/n\); for instance, \(n=5\) gives measure \(1/5\). As \(n\to\infty\), these measures tend to zero. More generally, for any fixed \(\varepsilon>0\), the transformed error set is \((0,1/n)\) when \(n>2\) and \(2>\varepsilon\), and is empty when \(2\le\varepsilon\). Its measure therefore tends to zero at every threshold.
Check: On the support of \(f_n\), the original difference is \(n\) and the truncated difference is \(2\); outside it both differences are zero. Thus \(|T_2(f_n)-T_2(f)|\le |f_n-f|\) pointwise, as the 1-Lipschitz estimate predicts.
Why Continuity Alone Does Not Give the Result
Continuity at each real number gives a tolerance that may depend on that number. When the functions take values in a range that moves farther and farther along the real line, those tolerances need not control the error sets uniformly. The following example shows that this is a genuine obstruction, even though the original sequence converges in measure.
Worked Example: A Continuous Transformation That Fails
Take \(X=\mathbb{R}\) with its Borel sigma-algebra and Lebesgue measure \(m\). Let \(f(x)=x\) and $$ f_n(x)=x+\frac1n\chi_{[n,n+1]}(x). $$ These are Borel measurable real-valued functions. For a fixed \(\varepsilon>0\), the original error set is \([n,n+1]\) if \(1/n>\varepsilon\), and is empty if \(1/n\le\varepsilon\). In the first case its measure is \(1\); in the second its measure is \(0\). Since \(1/n\le\varepsilon\) for all sufficiently large \(n\), the error-set measures tend to zero, so \(f_n\to f\) in measure.
Now use the continuous, but not uniformly continuous, function \(\Phi(t)=t^2\). Outside \([n,n+1]\), the transformed difference is zero. On that interval, $$ |\Phi(f_n(x))-\Phi(f(x))| =(x+1/n)^2-x^2 =\frac{2x}{n}+\frac1{n^2}. $$ For every \(x\in[n,n+1]\), this is at least \(2+1/n^2>1\). Thus the transformed error set at threshold \(1\) contains \([n,n+1]\), has measure \(1\), and does not tend to zero. In fact, it is exactly \([n,n+1]\), because the transformed difference vanishes outside this interval.
Check: The failure is consistent with non-uniform continuity: for \(s=n\) and \(t=n+1/n\), \(|s-t|=1/n\to0\), while \(|s^2-t^2|=2+1/n^2>2\). A small input difference does not ensure a small output difference uniformly over all real inputs.
Using the Result Correctly
For a general uniformly continuous \(\Phi\), do not assume a bound of the form \(|\Phi(s)-\Phi(t)|\le C|s-t|\). Uniform continuity only promises, for each output threshold, a suitable input threshold; the proof uses the resulting inclusion of error sets, not a linear estimate. Absolute value and truncation have the stronger 1-Lipschitz property, but that property is not required for the theorem.
When applying the theorem, check that the functions are measurable and real-valued, that the transformed function is uniformly continuous on all of \(\mathbb{R}\), and that convergence in measure is being tested at each fixed positive threshold. No finiteness assumption on \(\mu(X)\) is needed. If the transformation is only continuous, examine whether the input values can range over an unbounded set: the square-function example shows that continuity by itself cannot replace uniform continuity in this general statement.
Check Your Understanding
Use the error-set definition and identify the threshold used in each argument.
- For a uniformly continuous \(\Phi:\mathbb{R}\to\mathbb{R}\), explain why the transformed error set at threshold \(\varepsilon\) is contained in an original error set at some fixed positive threshold.
- On \([0,1]\), let \(f_n(x)=x-1/n\) and \(f(x)=x\). Compute the error-set measure for \(|f_n|\to|f|\) at a fixed threshold \(\varepsilon>0\), or give a bound that proves convergence in measure.
- For \(M>0\), verify directly from the definition of \(T_M\) that \(T_M\) is nondecreasing and that its increase between any \(s\le t\) is at most \(t-s\).
- In the spike example, why does the measure of the transformed error set tend to zero even though the truncated difference is \(2\) on its support?
- In the square-function example, compute the transformed error-set measure at threshold \(1\) and explain why this does not contradict the uniform-continuity theorem.