Uniform Convergence and Integration

Why Uniform Approximation Matters for Integration

An integral measures the accumulated values of a function across an interval. If two functions are uniformly close, their values differ only slightly at every point, so their integrals should also be close. Uniform convergence makes this control available for an entire sequence at once. It also has a deeper use: a uniform limit of Riemann integrable functions is Riemann integrable, even when the limiting function is not continuous.

Let \([a,b]\) be a closed bounded interval. Recall that a bounded function is Riemann integrable if its upper and lower Darboux sums can be made arbitrarily close by a suitable partition. We will use the Darboux criterion for Riemann integrability established earlier in the course. The key point is to compare the upper and lower sums of a uniform limit with those of one sufficiently close approximating function.

Definition: A sequence \(f_n:[a,b]\to\mathbb{R}\) converges uniformly to \(f\) if for every \(\varepsilon>0\), there is an \(N\) such that \(|f_n(x)-f(x)|<\varepsilon\) for every \(n\geq N\) and every \(x\in[a,b]\). The index \(N\) may depend on \(\varepsilon\), but not on \(x\).

The dependence on \(x\) is exactly what uniform convergence rules out. For integration, this matters because an estimate that is good at each fixed point need not be good across enough of the interval to control the total area.

Uniform Limits Preserve Riemann Integrability

Theorem (Uniform Limits of Riemann Integrable Functions Are Riemann Integrable): Suppose each \(f_n:[a,b]\to\mathbb{R}\) is Riemann integrable and \(f_n\) converges uniformly to \(f:[a,b]\to\mathbb{R}\). Then \(f\) is Riemann integrable.

Proof. If \(a=b\), the interval has length zero and the integral criterion is immediate. Suppose \(a<b\), and write \(L=b-a\). First, \(f\) is bounded. Choose \(N\) such that \(|f(x)-f_N(x)|<1\) for every \(x\in[a,b]\). Since \(f_N\) is Riemann integrable, it is bounded; take \(M\geq0\) with \(|f_N(x)|\leq M\) throughout the interval. Then \(|f(x)|\leq |f(x)-f_N(x)|+|f_N(x)|<1+M\), so \(f\) is bounded.

Now let \(\varepsilon>0\). By uniform convergence, choose \(N\) such that

|f(x)-f_N(x)|<\delta \qquad\text{for every }x\in[a,b], \qquad \delta=\frac{\varepsilon}{4L}.

Because \(f_N\) is Riemann integrable, the Darboux criterion gives a partition \(P\) of \([a,b]\) for which \(U(f_N,P)-L(f_N,P)<\varepsilon/2\). On each subinterval of \(P\), the pointwise inequalities \(f(x)<f_N(x)+\delta\) and \(f(x)>f_N(x)-\delta\) imply that the supremum of \(f\) is at most the supremum of \(f_N\) plus \(\delta\), and the infimum of \(f\) is at least the infimum of \(f_N\) minus \(\delta\). Summing these estimates over the subintervals, whose lengths add to \(L\), gives

U(f,P)\leq U(f_N,P)+\delta L, \qquad L(f,P)\geq L(f_N,P)-\delta L.

Consequently,

U(f,P)-L(f,P) \leq U(f_N,P)-L(f_N,P)+2\delta L <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.

Since \(f\) is bounded and this can be done for every \(\varepsilon>0\), the Darboux criterion shows that \(f\) is Riemann integrable. \(\square\)

The proof uses one approximating function \(f_N\) on the whole interval. Uniform closeness limits how much the upper and lower sums can change, while integrability of \(f_N\) provides a partition with a small gap. With pointwise convergence alone, no single index need control the error at every point, so this argument would not work.

Worked Example: A Uniform Limit with a Discontinuity

Define \(g:[0,1]\to\mathbb{R}\) by \(g(x)=0\) for \(0\leq x<1/2\) and \(g(x)=1\) for \(1/2\leq x\leq1\). This step function is Riemann integrable, although it is discontinuous at \(1/2\). For \(n\geq1\), set

f_n(x)=g(x)+\frac{x}{n}.

Each \(f_n\) is Riemann integrable: it is the sum of the Riemann integrable step function \(g\) and the continuous function \(x/n\). Moreover,

|f_n(x)-g(x)|=\frac{x}{n}\leq\frac{1}{n} \qquad (x\in[0,1]).

Thus \(f_n\) converges uniformly to \(g\). The theorem applies even though the limit is discontinuous. Its integral is \(\int_0^1 g(x)\,dx=1/2\), since changing the value at the single point \(1/2\) does not affect the integral. Directly, \(\int_0^1 f_n(x)\,dx=1/2+1/(2n)\), which approaches \(1/2\). This example shows why uniform limits of integrable functions should not be confused with uniform limits of continuous functions: the latter are continuous, but the former need not be.

How Uniform Error Controls Integral Error

Once the limit is known to be integrable, uniform convergence also gives convergence of the integrals. For continuous functions this is the content of the earlier theorem Integration of a Uniform Limit. The same conclusion holds for Riemann integrable functions, and the estimate follows directly from the order properties of the Riemann integral.

Theorem (Convergence of Integrals under Uniform Convergence): Suppose each \(f_n:[a,b]\to\mathbb{R}\) is Riemann integrable and \(f_n\) converges uniformly to \(f\). Then \(f\) is Riemann integrable and \(\int_a^b f_n(x)\,dx\to\int_a^b f(x)\,dx\).

Proof. The preceding theorem shows that \(f\) is Riemann integrable. If \(a=b\), all the integrals are zero. Suppose \(a<b\), and let \(\varepsilon>0\). Uniform convergence gives an \(N\) such that for \(n\geq N\), \(|f_n(x)-f(x)|<\varepsilon/(b-a)\) at every \(x\in[a,b]\). Therefore

-\frac{\varepsilon}{b-a}<f_n(x)-f(x)<\frac{\varepsilon}{b-a} \qquad (x\in[a,b]).

Integrating these inequalities and using linearity and monotonicity of the Riemann integral yields

-\varepsilon \leq \int_a^b f_n(x)\,dx-\int_a^b f(x)\,dx \leq \varepsilon.

Hence the integrals converge. The same argument gives the quantitative estimate \(\left|\int_a^b f_n-\int_a^b f\right|\leq (b-a)\sup_{x\in[a,b]}|f_n(x)-f(x)|\). Thus the interval length scales the uniform error. \(\square\)

Worked Example: Estimating the Integral Error

On \([0,1]\), let \(f_n(x)=x^2+\frac{\sin(nx)}{n+2}\), where \(n\geq1\), and let \(f(x)=x^2\). Since \(|\sin(nx)|\leq1\),

|f_n(x)-f(x)| \leq\frac{1}{n+2} \qquad (x\in[0,1]).

This proves uniform convergence. The limit integral is \(\int_0^1x^2\,dx=1/3\). Integrating the additional term gives

\int_0^1 f_n(x)\,dx =\frac13+\frac{1-\cos(n)}{n(n+2)}.

Indeed, an antiderivative of \(\sin(nx)\) is \(-\cos(nx)/n\), so its integral from \(0\) to \(1\) is \((1-\cos n)/n\). Since \(0\leq1-\cos n\leq2\),

\left|\int_0^1 f_n(x)\,dx-\frac13\right| =\frac{1-\cos(n)}{n(n+2)} \leq\frac{2}{n(n+2)} \longrightarrow0.

The general uniform error estimate also gives an upper bound of \(1/(n+2)\), because the interval has length \(1\). The direct computation is sharper here, but the estimate works even when the integral of the error term is difficult to calculate.

Why Pointwise Convergence Is Not Enough

Pointwise convergence controls the sequence at each fixed input, but the inputs where a function is large can change with \(n\). Such moving concentrations can contribute a fixed amount to the integral while disappearing at every fixed point. The next example makes the distinction explicit.

Worked Example: Pointwise Convergence with No Integral Convergence

For integers \(n\geq3\), define \(h_n:[0,1]\to\mathbb{R}\) by

h_n(x)=n\max\{0,1-n|x-2/n|\}.

Each \(h_n\) is continuous. It vanishes outside \([1/n,3/n]\), is zero at both endpoints of that interval, and has value \(n\) at \(x=2/n\). Its graph on that interval is a triangle with base \(2/n\) and height \(n\), so

\int_0^1h_n(x)\,dx=\frac12\cdot\frac{2}{n}\cdot n=1.

For any fixed \(x\in[0,1]\), if \(x=0\), then \(h_n(x)=0\) for every \(n\). If \(x>0\), choose \(n>3/x\); then \(3/n<x\), so \(x\) lies outside the support of \(h_n\) and \(h_n(x)=0\). Thus \(h_n(x)\to0\) at every fixed \(x\), and the pointwise limit is the zero function. But the integrals all equal \(1\), whereas the integral of the limit is \(0\). Convergence is not uniform: at \(x=2/n\), \(|h_n(x)-0|=n\), so the supremum error does not tend to zero.

Using the Results Carefully

Uniform convergence gives two useful conclusions on a closed bounded interval: it preserves Riemann integrability, and it ensures convergence of the integrals. The first conclusion is broader than the familiar fact that uniform limits of continuous functions are continuous. A Riemann integrable sequence may have a discontinuous uniform limit, as the step-function example showed.

The interval matters in the integral estimate. A uniform error of at most \(\delta\) contributes at most \((b-a)\delta\) to the absolute integral error. On a longer interval, the same pointwise error can produce a larger total error. The estimate also depends on having a finite interval; the argument does not automatically extend to an unbounded interval or an improper integral.

Finally, pointwise convergence should not be substituted for uniform convergence without additional hypotheses. The triangle functions converge pointwise to zero but retain integral \(1\), because their tall, narrow supports move toward the endpoint. Uniform convergence rules out this behavior by controlling the error simultaneously at every point.

Takeaway: On a closed bounded interval, a uniform limit of Riemann integrable functions is Riemann integrable, and uniform error bounds control the difference of the integrals. Pointwise convergence alone provides neither conclusion in general.

Check Your Understanding

Use the Darboux-sum argument and the integral estimates in this tutorial to answer the following questions.

Why does uniform closeness to one Riemann integrable function help make the upper and lower sums of the limit close?
In the integral error estimate, where does the factor \(b-a\) enter?
Can a uniform limit of Riemann integrable functions be discontinuous? Give the example from this tutorial.
Why do the functions \(h_n\) converge pointwise to zero even though their integrals remain equal to one?
Which hypothesis in the convergence-of-integrals theorem fails for the triangle functions?