Why Sums and Products Need Different Estimates
Convergence in measure controls the size of the set where an error exceeds a fixed threshold; it does not say that the errors are uniformly small at every point. For sums, the triangle inequality turns a large sum error into a large error in at least one summand. For products, the corresponding estimate includes the sizes of the factors. On a finite measure space, convergence in measure supplies the needed control of those sizes, even though it does not imply a uniform pointwise bound.
Throughout, unless stated otherwise, \((X,\mathcal{M},\mu)\) is a finite measure space: \(\mu(X)<\infty\). Functions are measurable and real-valued, so they are finite at every point. A sequence \(f_n\) converges in measure to \(f\) if, for every \(\varepsilon>0\),
The inequality in the error set is strict. Convergence in measure is not pointwise convergence or uniform convergence; the condition concerns the measures of error sets at each fixed threshold.
Sums: A Direct Error-Set Bound
Proof. Fix \(\varepsilon>0\). At any \(x\in X\), if both \(|f_n(x)-f(x)|\le\varepsilon/2\) and \(|g_n(x)-g(x)|\le\varepsilon/2\), the triangle inequality gives \(|f_n(x)+g_n(x)-f(x)-g(x)|\le\varepsilon\). Therefore
By subadditivity of the measure, the measure of the set on the left is at most the sum of the measures of the two sets on the right. Each of those measures tends to zero by the assumed convergence in measure, using the fixed threshold \(\varepsilon/2\). Their sum tends to zero, proving the assertion. \(\square\)
Worked Example: Adding Two Moving Indicators
On \([0,1]\) with Lebesgue measure \(m\), for integers \(n\ge2\) set \(f_n=\chi_{(0,1/n)}\) and \(g_n=\chi_{(1/2,\,1/2+1/n)}\). Both sequences converge in measure to the zero function: for any fixed \(\varepsilon>0\), if \(0<\varepsilon<1\), each error set is its indicated interval and has measure \(1/n\); if \(\varepsilon\ge1\), the error set is empty because the indicator values equal \(1\), not a value strictly greater than \(\varepsilon\). In either case its measure tends to zero.
For \(n\ge3\), the intervals are disjoint: \(1/n\le1/3<1/2\). Thus \(f_n+g_n\) equals \(1\) on their union and \(0\) elsewhere, so it converges in measure to zero. More explicitly, for \(0<\varepsilon<1\) its error set is the union, of measure \(1/n+1/n=2/n\); for \(\varepsilon\ge1\) the strict error set is empty. Check: at \(n=4\) and \(\varepsilon=1/2\), the two intervals each have length \(1/4\), are disjoint, and the error-set measure is \(1/4+1/4=1/2\), as the formula \(2/n\) gives.
Why Products Require Control of Factor Sizes
A useful intermediate fact is that a sequence converging in measure on a finite measure space cannot have large values on a substantial set, eventually. This is often called being bounded in measure. The bound here is about error-set measures, not a pointwise bound on the functions.
Proof. Since \(v\) is finite-valued, the sets \(\{|v|>K/2\}\), as \(K\) increases, decrease to the empty set. They are contained in \(X\), which has finite measure. Continuity of measure from above therefore gives \(\mu(\{|v|>K/2\})\to0\) as \(K\to\infty\). Choose \(K>0\) so that this measure is less than \(\eta/2\). For every \(n\), the triangle inequality gives
By convergence in measure at the fixed threshold \(K/2\), there is an \(N\) such that \(\mu(\{|v_n-v|>K/2\})<\eta/2\) whenever \(n\ge N\). Subadditivity now gives \(\mu(\{|v_n|>K\})<\eta\) for all such \(n\), as required. \(\square\)
Proof. Fix \(\varepsilon>0\) and \(\eta>0\). Since \(v\) is finite-valued, choose \(K>0\) with \(\mu(\{|v|>K/2\})<\eta/3\), using continuity from above as in the preceding proof. Convergence of \(v_n\) to \(v\) in measure gives an \(N_1\) such that \(\mu(\{|v_n-v|>K/2\})<\eta/3\) for \(n\ge N_1\). For these \(n\), the inclusion \[ \{|v_n|>K\}\subseteq\{|v|>K/2\}\cup\{|v_n-v|>K/2\} \] shows \(\mu(\{|v_n|>K\})<2\eta/3\). Since \(u_n\to0\) in measure, choose \(N_2\) such that \(\mu(\{|u_n|>\varepsilon/K\})<\eta/3\) for \(n\ge N_2\). If \(|u_nv_n|>\varepsilon\), then either \(|v_n|>K\), or \(|v_n|\le K\) and hence \(|u_n|>\varepsilon/K\). Thus for \(n\ge\max(N_1,N_2)\), the product error set has measure less than \(\eta\). Since \(\eta\) was arbitrary, its measure tends to zero. \(\square\)
Proof. The identity \[ f_ng_n-fg=(f_n-f)g_n+f(g_n-g) \] holds at every \(x\in X\). Apply the preceding product lemma with \(u_n=f_n-f\), which converges in measure to zero, and \(v_n=g_n\), which converges in measure to \(g\). It follows that \((f_n-f)g_n\to0\) in measure. Apply the same lemma with \(u_n=g_n-g\) and the constant sequence \(v_n=f\), which converges in measure to \(f\). It follows that \(f(g_n-g)\to0\) in measure. The sum theorem then shows that their sum converges in measure to zero. The displayed identity proves \(f_ng_n\to fg\) in measure. \(\square\)
Worked Example: Products of Unbounded Spikes
On \([0,1]\) with Lebesgue measure, let \(f_n=n\chi_{(0,1/n^2)}\). For a fixed \(\varepsilon>0\), if \(n\le\varepsilon\), then \(f_n\) never exceeds \(\varepsilon\), so the strict error set \(\{|f_n|>\varepsilon\}\) is empty. If \(n>\varepsilon\), that set is \((0,1/n^2)\), with measure \(1/n^2\). Consequently its measure tends to zero and \(f_n\to0\) in measure.
Now \(f_n^2=n^2\chi_{(0,1/n^2)}\). For fixed \(\varepsilon>0\), the error set \(\{|f_n^2|>\varepsilon\}\) is empty if \(n^2\le\varepsilon\), and otherwise is \((0,1/n^2)\), of measure \(1/n^2\). Its measure again tends to zero, so \(f_n^2\to0\) in measure, in agreement with the product theorem. Check: at \(n=4\), \(f_4\) has value \(4\) on an interval of length \(1/16\), while \(f_4^2\) has value \(16\) on that same interval. At threshold \(\varepsilon=1/2\), both error sets have measure \(1/16\). The pointwise heights grow, but their supporting sets shrink.
Worked Example: Products with Nonzero Limits
On \([0,1]\) with Lebesgue measure, define \(f_n=1+\chi_{(0,1/n)}\) and \(g_n=2+\chi_{(0,1/n)}\). Their respective limits in measure are the constant functions \(1\) and \(2\): for \(0<\varepsilon<1\), each error set has measure \(1/n\); for \(\varepsilon\ge1\), each strict error set is empty.
Writing \(I_n=\chi_{(0,1/n)}\), we have \(I_n^2=I_n\), and therefore \[ f_ng_n=(1+I_n)(2+I_n)=2+3I_n. \] The product limit is \(2\). For \(0<\varepsilon<3\), the error set \(\{|f_ng_n-2|>\varepsilon\}\) is \((0,1/n)\), of measure \(1/n\). For \(\varepsilon\ge3\), it is empty because the error on that interval is exactly \(3\) and the inequality is strict. Hence \(f_ng_n\to2\) in measure. Check: when \(n=4\) and \(\varepsilon=2\), the product differs from \(2\) by \(3\) on an interval of length \(1/4\), so the error-set measure is \(1/4\).
Why Finite Measure Matters for Products
The sum estimate works on any measure space, but the product proof uses finiteness to control the sets where the limiting factor is large. Without a finite measure assumption, convergence in measure of one factor need not make its product with a fixed finite-valued function converge in measure.
Worked Example: A Product Failure on the Real Line
Take \(\mathbb{R}\) with its Borel sigma-algebra and Lebesgue measure \(m\), which is an infinite measure space. Define \(g(x)=e^x\) and \(f_n(x)=e^{-x}\chi_{[n,n+1]}(x)\). These are finite-valued measurable functions. For each fixed \(\varepsilon>0\), choose \(N\) so that \(e^{-n}\le\varepsilon\) for \(n\ge N\). On \([n,n+1]\), \(0\le f_n(x)\le e^{-n}\); outside it, \(f_n(x)=0\). Thus for all \(n\ge N\), the strict error set \(\{|f_n|>\varepsilon\}\) is empty. In particular, \(f_n\to0\) in measure.
However, \(f_ng=\chi_{[n,n+1]}\). At threshold \(\varepsilon=1/2\), the product error set relative to the proposed limit \(0\) is exactly \([n,n+1]\), whose Lebesgue measure is \((n+1)-n=1\) for every \(n\). Therefore \(f_ng\) does not converge in measure to zero. Correction to the tempting argument “\(f_n\to0\) and \(g\) is finite-valued, so \(f_ng\to0\)”: finiteness at each point is not a uniform bound on \(g\), and on this infinite measure space the large values of \(g\) occur on sets that defeat the product estimate.
Common Pitfalls and a Practical Test
There are two distinct errors to avoid. First, convergence in measure does not mean that the functions are uniformly bounded pointwise: in the spike example, \(\sup_{x\in[0,1]}f_n(x)=n\), which grows without bound. The finite-measure product theorem does not require such a bound; it uses control of error-set measures instead. Second, the product theorem stated here is not an unrestricted assertion for every measure space. The real-line example shows that its finite-measure hypothesis cannot simply be deleted.
When checking a proposed sum or product limit, identify the error set at a fixed \(\varepsilon>0\). For sums, split the error threshold into two parts, \(\varepsilon/2\) and \(\varepsilon/2\), and use subadditivity. For products on a finite measure space, first control the set where a factor is large, then control the error in the other factor at the resulting fixed threshold. In every case, convergence in measure requires the error-set measure to tend to zero for each fixed positive threshold.
Check Your Understanding
For each question, state the measure space and use strict error sets of the form \(\{|h_n-h|>\varepsilon\}\).
- Prove that the sum of three sequences converging in measure on a measure space converges in measure to the sum of their limits.
- On \([0,1]\) with Lebesgue measure, let \(h_n=n\chi_{(0,1/n^3)}\). For fixed \(\varepsilon>0\), compute the error-set measure for all sufficiently large \(n\), and determine the limit in measure.
- Explain where finiteness of \(\mu(X)\) is used in the proof that a sequence converging in measure is bounded in measure.
- For \(f_n=1+\chi_{(0,1/n)}\) and \(g_n=2-\chi_{(0,1/n)}\) on \([0,1]\), compute \(f_ng_n\) and verify its limit in measure.
- In the real-line counterexample, compute the measure of \(\{|f_ng|>1/2\}\) and explain why it does not tend to zero.