Algebra and Almost-Everywhere Limits
When two sequences of real numbers converge, their sums and products converge to the corresponding sums and products of their limits. For sequences of functions, the same pointwise fact applies wherever both sequences converge. The measure-theoretic detail is that each sequence may have its own exceptional null set, so the pointwise argument must be applied outside their union.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; it is not assumed finite, sigma-finite, or complete. All functions in this tutorial are finite real-valued and \(\mathcal{M}\)-measurable, and all limits are finite real-valued and measurable. A set is null if it belongs to \(\mathcal{M}\) and has measure zero. No conclusion below requires the whole sequence to converge at every point.
We use the elementary fact that a finite union of measurable null sets is null: if \(\mu(N_1)=\mu(N_2)=0\), then \(N_1\cup N_2\in\mathcal{M}\) and \(\mu(N_1\cup N_2)\le\mu(N_1)+\mu(N_2)=0\). Also, sums and products of finite real-valued measurable functions are measurable, since addition and multiplication are continuous maps on \(\mathbb{R}^2\). Thus the algebraic combinations below are functions of the same measurable type.
The Algebraic Limit Theorem
Proof. By the two almost-everywhere convergence assumptions, there are measurable null sets \(N_f,N_g\) such that \(f_n(x)\to f(x)\) for every \(x\notin N_f\), and \(g_n(x)\to g(x)\) for every \(x\notin N_g\). Put \(N=N_f\cup N_g\). The finite-union estimate gives \(\mu(N)=0\). Fix any \(x\in X\setminus N\). Both real sequences then converge at this same \(x\).
For the sum, let \(\varepsilon>0\). There are indices \(n_f,n_g\) such that \(n\ge n_f\) implies \(|f_n(x)-f(x)|<\varepsilon/2\), and \(n\ge n_g\) implies \(|g_n(x)-g(x)|<\varepsilon/2\). For \(n\ge\max(n_f,n_g)\), the triangle inequality gives
So the sums converge at \(x\). For a fixed scalar \(c\), if \(c=0\), then \(cf_n(x)=cf(x)=0\) for every \(n\). If \(c\ne0\), choose an index after which \(|f_n(x)-f(x)|<\varepsilon/|c|\). Then \(|cf_n(x)-cf(x)|=|c|\,|f_n(x)-f(x)|<\varepsilon\). Thus the scalar multiples converge at \(x\).
For the product, convergence of \(g_n(x)\) to the finite real number \(g(x)\) implies that there is an index \(n_0\) such that \(n\ge n_0\) gives \(|g_n(x)-g(x)|<1\), and hence \(|g_n(x)|\le |g(x)|+1\). For any \(\varepsilon>0\), choose indices so that, for all sufficiently large \(n\),
Take \(n\) large enough for these bounds and the bound \(|g_n(x)|\le |g(x)|+1\) to hold together. The identity \(f_ng_n-fg=(f_n-f)g_n+f(g_n-g)\), evaluated at \(x\), yields
This proves product convergence at \(x\), including when \(f(x)=0\) or \(g(x)=0\), since the denominators used above are positive. As \(x\) was arbitrary outside the null set \(N\), all three asserted convergences hold \(\mu\)-a.e. \(\square\)
The proof is pointwise outside one common exceptional set. It does not require a uniform bound on the functions over \(X\): the product estimate uses only that, at each fixed point of convergence, a convergent real sequence is eventually bounded.
Worked Example: Sums, a Negative Scalar, and Products
Take \(X=[0,1]\), \(\mathcal{M}=\mathcal{L}\) restricted to \([0,1]\), and \(\mu=m\). For \(n\ge1\), set \(f_n(x)=x+1/n\) and \(g_n(x)=2-x\). These are finite measurable functions and converge everywhere to \(f(x)=x\) and \(g(x)=2-x\), respectively. The theorem gives the limits of their algebraic combinations.
For the sum, \((f_n+g_n)(x)=x+1/n+2-x=2+1/n\), so its limit is \(2=f(x)+g(x)\). For the scalar multiple, \((-2f_n)(x)=-2x-2/n\), which tends to \(-2x\). For the product, direct expansion gives
For an exact check at \(x=1/2\) and \(n=4\), \(f_4(1/2)=1/2+1/4=3/4\), and \(g_4(1/2)=2-1/2=3/2\). The sum is \(3/4+3/2=9/4=2+1/4\), the scalar multiple is \(-2(3/4)=-3/2\), and the product is \((3/4)(3/2)=9/8\). The corresponding limits at \(x=1/2\) are \(2\), \(-1\), and \((1/2)(3/2)=3/4\). In particular, the product formula gives \(3/4+(3/2)/4=3/4+3/8=9/8\) at this \(n\), agreeing with direct multiplication.
Check: the explicit remainder in the product is \((2-x)/n\). At \(x=1/2\), \(n=4\), it is \((3/2)/4=3/8\); adding it to the product limit \(3/4\) gives \(9/8\). Since \(0\le 2-x\le2\) on \([0,1]\), this remainder is at most \(2/n\), which tends to zero. This extra bound verifies the product limit uniformly in this particular example, although uniform convergence is not needed for the theorem.
Worked Example: Different Exceptional Sets
Let \(X=[0,1]\) with Lebesgue measure. Define, for \(n\ge1\),
These functions are measurable: each is given by measurable formulas on measurable pieces. The sequence \(f_n\) converges to \(x\) at every \(x>0\), and fails to converge at \(x=0\), where its values alternate between \(-1\) and \(1\). Thus it converges \(\mu\)-a.e. to the measurable function \(f(x)=x\). The sequence \(g_n\) converges to \(2x\) at every \(x<1\), and fails to converge at \(x=1\). It converges \(\mu\)-a.e. to \(g(x)=2x\). The exceptional sets \(\{0\}\) and \(\{1\}\) each have Lebesgue measure zero, so their union is also null.
For \(0<x<1\), the sum is \((x+1/n)+(2x-1/n)=3x\) for every \(n\), so it converges to \(3x=f(x)+g(x)\). On the same interval,
At \(x=1/2\), \(n=2\), the values are \(f_2=1/2+1/2=1\) and \(g_2=1-1/2=1/2\); their sum is \(3/2\) and product is \(1/2\). The limits there are \(f=1/2\), \(g=1\), whose sum is \(3/2\) and product is \(1/2\). The displayed product expansion also gives \(2(1/2)^2+(1/2)/2-1/2^2=1/2+1/4-1/4=1/2\). At the endpoints, the sequences need not converge; those points belong to the allowed exceptional set. The result is almost-everywhere convergence, not everywhere convergence.
Worked Example: A Product on an Unbounded Domain
Let \(X=(0,\infty)\) with its Borel sigma-algebra and Lebesgue measure. For \(n\ge1\), define \(f_n(x)=x+1/n\) and \(g_n(x)=1/(1+x)+1/n\). Both sequences converge everywhere on \(X\), to \(f(x)=x\) and \(g(x)=1/(1+x)\). Their product satisfies the exact identity
For each fixed \(x>0\), the last three terms tend to zero, so the product tends to \(x/(1+x)=f(x)g(x)\). For a numerical check at \(x=2\), \(n=4\), \(f_4=2+1/4=9/4\) and \(g_4=1/3+1/4=7/12\). Their product is \(63/48=21/16\). The formula gives \(2/3+2/4+1/(4\cdot3)+1/16=2/3+1/2+1/12+1/16=21/16\). Its difference from the limit \(2/3\) is \(21/16-2/3=31/48\), which equals \(2/4+1/12+1/16=31/48\), the sum of the three remainder terms.
The domain is unbounded and the term \(x/n\) is not bounded uniformly over all \(x>0\) for fixed \(n\). Nevertheless, for each fixed \(x\), \(x/n\to0\); the theorem concerns pointwise convergence outside a null set and makes no claim of uniform convergence on this domain.
A Product Estimate to Keep Straight
A common error is to assume that product differences are controlled by the sum of the two factor differences alone. For real numbers, the proposed inequality \(|ab-uv|\le |a-u|+|b-v|\) is false in general. For a concrete failure, take \(a=b=2\) and \(u=v=1\). Then the left side is \(|4-1|=3\), whereas the right side is \(|2-1|+|2-1|=2\).
The same mistake can arise when checking convergent sequences: \(f_n=g_n=1+1/n\), with limits \(f=g=1\), gives at \(n=1\) a product error \(|2\cdot2-1\cdot1|=3\), but the sum of factor errors is \(1+1=2\). The correction is the factorization used in the proof:
The extra factor \(|g_n|\) matters. At \(n=1\) in this example, the corrected right side is \(1\cdot2+1\cdot1=3\), which is valid. In a convergence proof, \(g_n\) is eventually bounded at each point where it converges, and that boundedness combines with the two factor limits to make the right side tend to zero.
The theorem also covers any finite number of sums by applying the two-sequence result repeatedly. At each application only finitely many exceptional null sets are combined, and a finite union of null sets remains null. It does not assert a corresponding result for infinite sums: infinitely many terms require additional hypotheses and a separate argument.
Check Your Understanding
For each question, distinguish what holds outside a null set from what holds at every point.
- If \(f_n\to f\) \(\mu\)-a.e. and \(g_n\to g\) \(\mu\)-a.e., which measurable null set should be removed to apply pointwise limit rules to both sequences at once?
- On \([0,1]\) with Lebesgue measure, let \(f_n(x)=x+1/n\) and \(g_n(x)=3x\). Find the pointwise limits of \(f_n+g_n\), \(-f_n\), and \(f_ng_n\).
- Why does convergence \(g_n(x)\to g(x)\) at a fixed point imply that \(|g_n(x)|\) is eventually bounded? Give a bound that works for all sufficiently large \(n\).
- For the two sequences with exceptional sets \(\{0\}\) and \(\{1\}\) in the second worked example, identify the common exceptional set and its Lebesgue measure.
- Check the proposed inequality \(|ab-uv|\le |a-u|+|b-v|\) using \(a=b=3\) and \(u=v=1\). State the factorization that gives a valid product estimate instead.