What Does Uniqueness Mean?
A sequence can approach a limit in different senses, and the meaning of “unique” depends on the sense being used. For convergence in measure, values on a null set cannot be detected: changing a candidate limit there leaves every error-set measure unchanged. The correct uniqueness statement is therefore equality almost everywhere, rather than equality at every point.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space, with no assumption that \(\mu(X)\) is finite or that the space is sigma-finite or complete. The functions in this tutorial are measurable, finite-valued functions \(X\to\mathbb{C}\); real-valued functions are included. Convergence in measure is defined by the measures of strict threshold error sets:
The threshold \(t\) is fixed while \(n\) tends to infinity. The definition does not ask for pointwise convergence, and it does not require the total measure of \(X\) to be finite. We will show that if the same sequence converges in measure to two functions, then the two functions can differ only on a null set.
The Threshold-Set Argument
The triangle inequality connects the error from \(f_n\) to \(f\), the error from \(f_n\) to \(g\), and the difference between the proposed limits. To use it, first fix a threshold for the difference between \(f\) and \(g\). If that difference is larger than the threshold, at least one of the two errors must exceed a suitably smaller threshold.
Proof. Fix \(t>0\), and define the measurable sets \(E_t=\{x:|f(x)-g(x)|>t\}\), \(A_n=\{x:|f_n(x)-f(x)|>t/2\}\), and \(B_n=\{x:|f_n(x)-g(x)|>t/2\}\). For each \(x\in E_t\), the triangle inequality gives \[ |f(x)-g(x)|\le |f(x)-f_n(x)|+|f_n(x)-g(x)|. \] If \(x\) were outside both \(A_n\) and \(B_n\), each term on the right would be at most \(t/2\), so \(|f(x)-g(x)|\le t\), contrary to \(x\in E_t\). Thus \(E_t\subseteq A_n\cup B_n\) for every \(n\). By monotonicity and subadditivity of measure, \[ 0\le\mu(E_t)\le\mu(A_n)+\mu(B_n). \] The two convergence-in-measure assumptions imply that \(\mu(A_n)\to0\) and \(\mu(B_n)\to0\), since \(t/2>0\) is fixed. The right side therefore tends to zero, so \(\mu(E_t)=0\). This conclusion holds for every \(t>0\). Every point where \(f\ne g\) has \(|f(x)-g(x)|>1/k\) for some positive integer \(k\), and hence \[ \{x:f(x)\ne g(x)\}=\bigcup_{k=1}^{\infty} E_{1/k}. \] A countable union of null sets is null. Therefore \(f=g\) \(\mu\)-almost everywhere. \(\square\)
The proof handles sets of infinite measure as well as finite ones. Indeed, the bound shows that the fixed set \(E_t\) has measure at most a quantity tending to zero. It follows that \(\mu(E_t)=0\), whether its measure was initially finite or potentially infinite. No subtraction of infinite quantities is involved.
Examples: Checking What the Result Says
Worked Example: A Constant Sequence Cannot Have Two Distinct Limits
On \([0,1]\) with Lebesgue measure \(m\), let \(f_n(x)=1/n\) for every \(x\in[0,1]\). Verify that \(f_n\to0\) in measure, and test the different candidate \(g(x)=1\).
For any fixed \(t>0\), the error set for the zero function is \[ \{x:|f_n(x)-0|>t\}= \begin{cases} [0,1],&1/n>t,\\ \varnothing,&1/n\le t. \end{cases} \] Once \(n\ge 1/t\), the second case applies, so its measure is zero for all sufficiently large \(n\). This proves convergence in measure to zero. For the candidate \(g=1\), take the fixed threshold \(t=1/4\). For every \(n\ge2\), \(1-1/n\ge1/2>1/4\). Thus \(\{|f_n-g|>1/4\}=[0,1]\) for every \(n\ge2\), and its measure is \(1\), not a quantity tending to zero. Check: the two proposed limits differ on all of \([0,1]\), a set of measure \(1\), and the second candidate fails the definition exactly as uniqueness predicts.
Worked Example: Two Pointwise Different Representatives of One Limit
On \([0,1]\) with Lebesgue measure, set \(f_n(x)=x/n\), \(f(x)=0\), and define \(g(0)=1\) and \(g(x)=0\) for \(0<x\le1\). Check directly that the sequence converges in measure to both \(f\) and \(g\).
Fix \(t>0\). For \(f\), the error set is \(\{x\in[0,1]:x/n>t\}\). If \(nt<1\), this is \((nt,1]\), of measure \(1-nt\); if \(nt\ge1\), it is empty, of measure zero. In particular, for all sufficiently large \(n\), \(nt\ge1\), so the error-set measures tend to zero. For \(0<t<1\), the error set for \(g\) consists of the same interval \((nt,1]\), when \(nt<1\), together with the point \(0\), where \(|f_n(0)-g(0)|=1>t\). Its measure is still \(1-nt\), because a singleton has Lebesgue measure zero. When \(nt\ge1\), only the singleton can contribute, so the measure is zero. For \(t\ge1\), the point \(0\) does not belong to the error set because the inequality in the definition is strict, and the other points have error \(x/n\le1/n\); the error-set measure again tends to zero.
Check: at \(t=1/4\), the error-set measure for either limit is \(1-n/4\) when \(n<4\), and zero when \(n\ge4\); at \(n=4\), the endpoint \(x=1\) has error exactly \(1/4\), so it is excluded by the strict inequality. The two limits disagree only at \(x=0\), a null set. This is consistent with uniqueness almost everywhere, but shows why pointwise uniqueness would be false.
Worked Example: Uniqueness on an Infinite Measure Space
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with counting measure, so every point has measure \(1\) and \(\mu(X)=\infty\). Set \(f_n(k)=1/n\) for every \(k\in\mathbb{N}\). Verify convergence in measure to zero and see why the constant function \(g(k)=1\) is not another limit.
For any fixed \(t>0\), if \(1/n\le t\), then \(\{k:|f_n(k)|>t\}=\varnothing\). This holds for all sufficiently large \(n\), so the measures tend to zero even though the underlying space has infinite measure. For \(g=1\), use \(t=1/2\). For every \(n\ge3\), \(1-1/n\ge2/3>1/2\), so \(\{k:|f_n(k)-g(k)|>1/2\}=\mathbb{N}\), whose counting measure is infinite. Hence convergence to \(g\) fails. Check: at each sufficiently large index, the error-set measure for zero is \(0\), whereas the error-set measure for \(g\) is \(+\infty\). The uniqueness theorem applies without a finite-measure hypothesis.
Limits as Almost-Everywhere Equivalence Classes
The theorem has a converse worth stating: if two measurable functions agree almost everywhere, then a sequence converges in measure to one exactly when it converges in measure to the other. This explains precisely the sense in which the limit is unique.
Proof. Let \(N=\{x:f(x)\ne g(x)\}\), which is measurable and null by assumption. At every \(x\notin N\), \(|f_n(x)-f(x)|=|f_n(x)-g(x)|\). Therefore the two error sets \(\{|f_n-f|>t\}\) and \(\{|f_n-g|>t\}\) agree outside \(N\). Their symmetric difference is a subset of \(N\), and so is null. Two measurable sets whose symmetric difference is null have the same measure: each is contained in the other together with a null set, giving both measure inequalities by subadditivity. Their error-set measures are therefore equal for every \(n\) and \(t>0\). One sequence of these measures tends to zero exactly when the other does, proving both implications. \(\square\)
Combining the theorem and corollary, any sequence that has a limit in measure has exactly one such limit when functions equal \(\mu\)-almost everywhere are identified. This does not assert that the sequence has a limit: existence is a separate question. It also does not make convergence in measure the same as pointwise or almost-everywhere convergence; those are distinct modes of convergence.
A Threshold Pitfall
A frequent error is to use the same threshold on both error sets as on the difference of the limits. The triangle inequality does not give that set inclusion. For instance, on a one-point space, take the three values \(f=0\), \(g=1.2\), and \(f_n=0.6\), and choose \(t=1\). Then \(|f-g|=1.2>1\), while both \(|f_n-f|=0.6\) and \(|f_n-g|=0.6\) are at most \(1\). Thus the set where \(|f-g|>1\) is not contained in the union of the two sets where the errors exceed \(1\).
The needed thresholds need not be equal, but their sum must be at most the threshold for the difference. In detail, if \(\alpha,\beta>0\) and \(\alpha+\beta\le t\), then \[ \{|f-g|>t\}\subseteq \{|f_n-f|>\alpha\}\cup\{|f_n-g|>\beta\}. \] Indeed, outside the union both errors are at most \(\alpha\) and \(\beta\), so the triangle inequality gives \(|f-g|\le\alpha+\beta\le t\), contradicting \(|f-g|>t\). The symmetric choice \(\alpha=\beta=t/2\) used in the proof is convenient; for example, \(\alpha=t/3\) and \(\beta=2t/3\) work as well. The point is the sum condition, not a requirement that the thresholds be equal.
Check Your Understanding
For each question, pay attention to the fixed threshold and to whether the conclusion is pointwise or almost everywhere.
- In the uniqueness proof, why does \(\{f\ne g\}=\bigcup_{k\ge1}\{|f-g|>1/k\}\) hold?
- Suppose the error thresholds are \(\alpha>0\) and \(\beta>0\). What condition on \(\alpha,\beta\) ensures the threshold-set inclusion for \(\{|f-g|>t\}\)?
- Can a sequence converge in measure to two functions that differ at one point in \([0,1]\) with Lebesgue measure? Explain why or why not.
- Which part of the uniqueness proof would require modification if the error sets were defined using \(\ge t\) instead of \(>t\)?
- Does the uniqueness theorem require \(\mu(X)<\infty\)? Identify where the proof uses the measure-space assumptions.