Uniform Error and Integral Error
Uniform convergence controls the largest pointwise error across the whole space. To turn that control into an estimate for integrals, one also needs to know how much measure the space has. On a finite-measure space, an error bounded everywhere by \(\varepsilon\) contributes at most \(\varepsilon\mu(X)\) to the integral. On an infinite-measure space, even a very small uniform error may accumulate over a sufficiently large region.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space, not assumed complete or \(\sigma\)-finite unless stated. Functions are measurable and real-valued or complex-valued. A function \(h\) is integrable if \(\int_X |h|\,d\mu<\infty\); its integral is the Lebesgue integral. Uniform convergence \(f_n\to f\) means that for every \(\varepsilon>0\), there is an \(N\) such that \(|f_n(x)-f(x)|<\varepsilon\) for every \(x\in X\) and every \(n\ge N\). This is a bound at every point, not merely almost everywhere.
The Finite-Measure Estimate
Proof. Since \(f\) and \(g\) are integrable, \(f-g\) is integrable and linearity gives \(\int_X f\,d\mu-\int_Xg\,d\mu=\int_X(f-g)\,d\mu\). The triangle inequality for the Lebesgue integral gives \(\left|\int_X(f-g)\,d\mu\right|\le\int_X|f-g|\,d\mu\). By the definition of the supremum, \(|f(x)-g(x)|\le\|f-g\|_u\) for every \(x\in X\). Monotonicity of the integral therefore gives \(\int_X|f-g|\,d\mu\le\int_X\|f-g\|_u\,d\mu=\|f-g\|_u\mu(X)\). If \(\mu(X)=0\), this bound says both integrals are equal to zero; if \(X\) is empty, the integrals and the stated bound are also zero. \(\square\)
Proof. Apply the theorem to \(f_n\) and \(f\). For every \(n\), $$ \left|\int_X f_n\,d\mu-\int_X f\,d\mu\right| \le \mu(X)\|f_n-f\|_u. $$ The factor \(\mu(X)\) is finite and fixed, while \(\|f_n-f\|_u\to0\). Hence the difference of the integrals tends to zero. If \(\mu(X)=0\), every integral in the conclusion is zero, so the result also holds in that case. \(\square\)
There is also a direct connection with the Dominated Convergence Theorem, whose hypotheses include pointwise almost-everywhere convergence and domination by one integrable function. The finite-measure uniform-convergence result is a corollary of that theorem: uniform convergence supplies a useful integrable dominator for the tail. This argument also explains why it is harmless if a finite number of early functions are not integrable, provided the limit is integrable.
Indeed, suppose \(f_n\to f\) uniformly and \(f\) is integrable. Choose \(N\) such that \(\|f_n-f\|_u\le1\) for \(n\ge N\). Then \(|f_n|\le |f|+1\) for all \(n\ge N\), and \(g=|f|+1\) is integrable because \(\int_X g\,d\mu=\int_X|f|\,d\mu+\mu(X)<\infty\). The tail converges pointwise to \(f\) and is dominated by \(g\), so Dominated Convergence applies to that tail. Alternatively, the finite-measure estimate gives the stronger quantitative bound \(\left|\int f_n-\int f\right|\le\mu(X)\|f_n-f\|_u\) directly. If integrals of every \(f_n\) are to be considered, assume the early functions are integrable as well.
Worked Applications
Worked Example: A Uniform Approximation on a Finite Interval
Let \(X=[0,2]\), let \(\mathcal{M}\) be the Lebesgue sigma-algebra, and let \(\mu=m\) be Lebesgue measure restricted to \(X\). Set \(f(x)=x^2\) and \(f_n(x)=x^2+1/n\) for integers \(n\ge1\). Compute the integrals and verify the uniform estimate.
Both functions are measurable and integrable on \([0,2]\). For every \(x\in[0,2]\), \(|f_n(x)-f(x)|=1/n\), so \(\|f_n-f\|_u=1/n\) and \(m(X)=2\). Direct integration gives $$ \int_0^2 f(x)\,dx=\int_0^2x^2\,dx=\left[\frac{x^3}{3}\right]_0^2=\frac83 $$ and $$ \int_0^2 f_n(x)\,dx=\int_0^2x^2\,dx+\int_0^2\frac1n\,dx =\frac83+\frac2n. $$ Thus the integral error is exactly \(2/n\), and the theorem's bound gives \(m(X)\|f_n-f\|_u=2\cdot(1/n)=2/n\). Check: for \(n=4\), the integral is \(8/3+1/2=19/6\), whose difference from \(8/3=16/6\) is \(3/6=1/2=2/4\), as required.
Worked Example: A Finite Space with Unequal Point Masses
Let \(X=\{a,b,c\}\), \(\mathcal{M}\) be the set of all subsets of \(X\), and define \(\mu(\{a\})=2\), \(\mu(\{b\})=1\), and \(\mu(\{c\})=3\). Define \(f(a)=1\), \(f(b)=-2\), \(f(c)=0\), and \(f_n(a)=1+1/n\), \(f_n(b)=-2+(-1)^n/(2n)\), \(f_n(c)=0\). Compute both integrals and check convergence.
All functions on this finite measurable space are measurable and integrable. The pointwise errors have magnitudes \(1/n\), \(1/(2n)\), and \(0\), respectively, so \(\|f_n-f\|_u=1/n\). The limit integral is $$ \int_X f\,d\mu=2(1)+1(-2)+3(0)=0. $$ For each \(n\), $$ \int_X f_n\,d\mu =2\left(1+\frac1n\right)+\left(-2+\frac{(-1)^n}{2n}\right)+3(0) =\frac{4+(-1)^n}{2n}. $$ This tends to zero, the integral of \(f\). The total measure is \(2+1+3=6\), so the theorem bounds the absolute integral error by \(6/n\). Check: for \(n=2\), the values are \(f_2(a)=3/2\), \(f_2(b)=-7/4\), and \(f_2(c)=0\). The integral is \(2(3/2)+1(-7/4)=3-7/4=5/4\), which agrees with \((4+1)/(2\cdot2)=5/4\) and is at most \(6/2=3\).
Why Finite Measure Matters
Worked Example: Uniform Convergence with Diverging Integrals
Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), and \(\mu=m\), Lebesgue measure. For each integer \(n\ge1\), define \(f_n(x)=\frac1n\chi_{[0,n^2]}(x)\), and let \(f(x)=0\). Show that convergence is uniform but the integrals do not converge to the integral of the limit.
At every \(x\), the difference is either zero or \(1/n\), and it equals \(1/n\) for \(x\in[0,n^2]\). Since this interval is nonempty, \(\|f_n-f\|_u=1/n\to0\). Thus convergence is uniform. Each \(f_n\) is nonnegative and integrable, since $$ \int_{\mathbb{R}} f_n\,dm=\frac1n\,m([0,n^2])=\frac1n\cdot n^2=n. $$ The limit \(f=0\) is integrable with integral zero, whereas the integrals \(n\) do not converge to zero. The finite-measure estimate cannot be applied because \(m(\mathbb{R})=\infty\). Check: at \(n=3\), the support has measure \(9\), the function's height is \(1/3\), and the integral is \(9(1/3)=3\). If a single integrable function \(g\) dominated every \(f_n\), monotonicity would imply \(n=\int f_n\,dm\le\int g\,dm\) for every \(n\), which is impossible for finite \(\int g\,dm\).
The example shows that the smallness of the uniform error is not enough on an infinite-measure space: the region where the error is nonzero can grow so quickly that the total integral grows instead of shrinking. An additional hypothesis can still give convergence there—for example, an integrable dominating function—but uniform convergence by itself supplies no such control.
Common Mistakes and a Missing Hypothesis
Omitting the measure factor. The correct estimate is \(\left|\int_X(f-g)\,d\mu\right|\le\mu(X)\|f-g\|_u\), not merely \(\left|\int_X(f-g)\,d\mu\right|\le\|f-g\|_u\). For instance, on \([0,2]\) with Lebesgue measure, let \(f(x)=1/2\) and \(g(x)=0\). Then the uniform error is \(1/2\), but the integral error is \(\int_0^2(1/2)\,dx=1\). The factor \(m([0,2])=2\) gives the correct bound \(2(1/2)=1\).
Replacing uniform convergence by pointwise convergence. On \([0,1]\), set \(u_n=n\chi_{(0,1/n)}\) and \(u=0\). For each fixed \(x\in[0,1]\), \(u_n(x)\to0\): at \(x=0\) every value is zero, and for \(x>0\), \(1/n\le x\) eventually, so \(x\notin(0,1/n)\). But \(\int_0^1u_n\,dm=n(1/n)=1\) for every \(n\). The convergence is not uniform because \(\|u_n-u\|_u=n\), not a quantity tending to zero. Thus pointwise convergence, even everywhere, cannot replace the uniform hypothesis in this corollary.
Uniform convergence is stronger than pointwise almost-everywhere convergence: it requires the same error bound at every point once the index is large enough. On a finite-measure space, that uniform bound translates into integral control through the exact factor \(\mu(X)\). On an infinite-measure space, the worked counterexample shows why no analogous conclusion follows without further assumptions.
Check Your Understanding
Use the finite-measure estimate and its hypotheses to analyze each question.
- On a measure space with \(\mu(X)=5\), if \(\|f_n-f\|_u\le 1/(3n)\), what upper bound does the theorem give for \(\left|\int f_n\,d\mu-\int f\,d\mu\right|\)?
- Why is \(|f|+1\) integrable when \(f\) is integrable and \(\mu(X)<\infty\)?
- For \(f_n=(1/n)\chi_{[0,n^2]}\) on \(\mathbb{R}\), verify separately that \(\|f_n\|_u\to0\) and \(\int f_n\,dm\) does not tend to zero.
- On \([0,1]\), why does \(n\chi_{(0,1/n)}\to0\) pointwise but not uniformly? Compute each integral.
- If \(\mu(X)=0\), what are the integrals of integrable functions on \(X\), and does the finite-measure conclusion still hold?