Interchanging Limits and Integrals

Beyond Uniform Convergence

The previous tutorial established that uniform convergence lets us pass a limit through a Riemann integral on a closed bounded interval. The uniform error bound controls the accumulated error across the entire interval. But some useful sequences do not converge uniformly: their error may remain large close to one or a few points, even as it becomes small everywhere else.

A useful way to handle such sequences is to separate the interval into a large part where convergence is uniform and a small collection of exceptional intervals where it may not be. If the functions are also bounded by one common constant, the exceptional intervals contribute little to the integral because their total length is small. We develop this argument for Riemann integrable functions.

Definition: Let $f_n:[a,b]\to\mathbb{R}$ converge pointwise to $f$. We say that the convergence is uniform outside small exceptional intervals if, for every $\delta>0$, there are finitely many closed subintervals of $[a,b]$, with total length less than $\delta$, such that $f_n\to f$ uniformly on their complement in $[a,b]$.

The exceptional intervals may depend on $\delta$, but once they are chosen, the convergence on the complement is uniform. This condition does not require the functions to converge uniformly near the exceptional intervals. The common bound in the theorem below will control the error there instead.

A Criterion for Interchanging Limits and Integrals

Theorem (Interchanging Limits and Riemann Integrals Outside Small Intervals): Suppose $f_n:[a,b]\to\mathbb{R}$ are Riemann integrable and converge pointwise to a Riemann integrable function $f$. Suppose there is a constant $M\geq0$ such that $|f_n(x)|\leq M$ for every $n$ and $x\in[a,b]$. If the convergence is uniform outside small exceptional intervals, then $$ \lim_{n\to\infty}\int_a^b f_n(x)\,dx=\int_a^b f(x)\,dx. $$

Proof. If $a=b$, all the integrals are zero, so the conclusion holds. Assume $a<b$, and write $L=b-a$. Taking pointwise limits in $|f_n(x)|\leq M$ shows that $|f(x)|\leq M$ for every $x$. Thus $$ |f_n(x)-f(x)|\leq 2M \qquad (x\in[a,b]). $$ If $M=0$, both $f_n$ and $f$ are identically zero, and there is nothing to prove. Suppose $M>0$.

Let $\varepsilon>0$. Choose $\delta>0$ so small that $2M\delta<\varepsilon/2$. By the assumed convergence property, there are finitely many closed exceptional intervals whose total length is less than $\delta$, and convergence is uniform on their complement. Choose $n$ sufficiently large that on that complement, $$ |f_n(x)-f(x)|<\frac{\varepsilon}{2L}. $$ The complement consists of finitely many intervals, possibly with some endpoints omitted; endpoints do not affect Riemann integrals. Split the integral of $f_n-f$ over the exceptional intervals and the complementary intervals. On the exceptional intervals, the absolute error is at most $2M$, so their total contribution in absolute value is at most $2M\delta$. On the complement, the uniform error bound gives a contribution in absolute value at most $\frac{\varepsilon}{2L}L=\varepsilon/2$. Using the triangle inequality for these finitely many integrals, we obtain $$ \left|\int_a^b f_n(x)\,dx-\int_a^b f(x)\,dx\right| \leq 2M\delta+\frac{\varepsilon}{2} <\varepsilon. $$ This holds for all sufficiently large $n$, which proves the claimed convergence. $\square$

The proof uses two different estimates for two different parts of the interval. On the complement of the exceptional intervals, uniform convergence makes the error small pointwise. On the exceptional intervals, the common bound limits the error, while their small total length limits their contribution to the integral.

When Only Finitely Many Points Cause Trouble

A frequent special case is that convergence is uniform away from finitely many points. Small intervals around those points can be chosen to have arbitrarily small total length. This gives a convenient version of the criterion.

Corollary (Finitely Many Exceptional Points): Suppose $f_n$ are Riemann integrable on $[a,b]$, converge pointwise to a Riemann integrable function $f$, and satisfy $|f_n(x)|\leq M$ for a common finite constant $M$. If $f_n\to f$ uniformly on every closed set obtained by removing neighborhoods of finitely many points, where the neighborhoods can be chosen with arbitrarily small total length, then $$ \lim_{n\to\infty}\int_a^b f_n(x)\,dx=\int_a^b f(x)\,dx. $$

Proof. Given any $\delta>0$, choose neighborhoods of the exceptional points whose total length is less than $\delta$. Their closures can be taken to be finitely many closed intervals, merging any that overlap. By the stated uniform convergence away from these neighborhoods, convergence is uniform on their complement. Thus the hypotheses of the preceding theorem hold, and its conclusion follows. $\square$

Worked Example: A Limit with One Exceptional Endpoint

On $[0,1]$, define $$ f_n(x)=\frac{x^2}{x^2+1/n}. $$ For $x>0$, the denominator approaches $x^2$, so $f_n(x)\to1$. At $x=0$, $f_n(0)=0$ for every $n$. The pointwise limit is therefore $f(0)=0$ and $f(x)=1$ for $0<x\leq1$. This bounded step-type function is Riemann integrable, with $\int_0^1 f(x)\,dx=1$. Also, $0\leq f_n(x)\leq1$ everywhere.

Fix $\delta>0$. On $[\delta,1]$, the difference from the limit is $$ |f_n(x)-1|=\frac{1/n}{x^2+1/n}\leq\frac{1}{n\delta^2}. $$ The right side tends to zero, so convergence is uniform on $[\delta,1]$. The remaining interval $[0,\delta]$ has length $\delta$, which can be made arbitrarily small. The theorem applies and gives $$ \lim_{n\to\infty}\int_0^1 f_n(x)\,dx=1. $$ This conclusion does not rely on uniform convergence near $0$; it relies on the common bound and the short length of the interval where uniform convergence is not being used.

Worked Example: A Narrow Peak at an Interior Point

Let $c=1/3$, and define on $[0,1]$ $$ g_n(x)=\frac{1}{1+n(x-c)^2}. $$ At $x=c$, $g_n(c)=1$ for every $n$. At every $x\neq c$, the denominator tends to infinity, so $g_n(x)\to0$. The limit $g$ equals $1$ at $c$ and $0$ elsewhere. A function that is zero except at one point is Riemann integrable and has integral zero. Moreover, $0\leq g_n(x)\leq1$ on the whole interval.

Choose a small $\delta>0$ so that $[c-\delta,c+\delta]\subseteq[0,1]$. Outside this interval, $|x-c|\geq\delta$, and hence $$ |g_n(x)-g(x)|=g_n(x)\leq\frac{1}{1+n\delta^2}. $$ This upper bound tends to zero, proving uniform convergence on the complement of the exceptional interval. The interval has length $2\delta$, which can be made arbitrarily small. Therefore the theorem yields $$ \lim_{n\to\infty}\int_0^1 g_n(x)\,dx=0=\int_0^1 g(x)\,dx. $$ The functions do not converge uniformly on $[0,1]$: for $n\geq3$, the points $x_n=c+1/\sqrt n$ lie in $[0,1]$, and $|g_n(x_n)-g(x_n)|=1/2$. The theorem handles this persistent error because it is confined to intervals of arbitrarily small length.

Worked Example: Convergence to a Jump Function

For $n\geq1$, define $$ h_n(x)=\frac{n(x-\frac12)}{1+n|x-\frac12|} \qquad (x\in[0,1]). $$ Each $h_n$ is continuous, and $|h_n(x)|\leq1$. At points $x<1/2$, the numerator is negative and the absolute value of $h_n(x)$ tends to $1$, so $h_n(x)\to-1$. At points $x>1/2$, $h_n(x)\to1$, while $h_n(1/2)=0$. The pointwise limit is the Riemann integrable function that is $-1$ to the left of $1/2$, $1$ to the right, and $0$ at $1/2$.

Outside $[1/2-\delta,1/2+\delta]$, put $t=|x-1/2|$, so $t\geq\delta$. The difference between $h_n(x)$ and its limiting value has absolute value $$ \frac{1}{1+nt}\leq\frac{1}{1+n\delta}. $$ Thus the convergence is uniform outside that interval. The exceptional interval has length $2\delta$, and the sequence has a common bound. The theorem therefore permits the limit to pass through the integral. In this case, each integral is zero: reflection about $1/2$ gives $h_n(1-x)=-h_n(x)$, so the contributions on the two halves cancel. The limit function also has integral zero, in agreement with the theorem.

What the Criterion Does—and Does Not—Say

The common bound and the control of the exceptional intervals do separate work. A uniform bound alone does not show that the error is small where convergence is not uniform; it only ensures that the error cannot be arbitrarily large there. The intervals must also have small total length. Conversely, small intervals are not enough if the functions can become unbounded on them, since a large height can offset a small width.

The condition that the limit be Riemann integrable matters as well: it ensures that the integral on the right side exists in the Riemann sense. The theorem is a sufficient condition for interchanging the limit and integral, not a claim that every pointwise-convergent sequence can be treated this way. As the previous tutorial’s moving triangle example showed, pointwise convergence can coexist with integrals that do not converge to the integral of the pointwise limit.

Uniform convergence from the previous tutorial is the special case with no exceptional intervals: the error is controlled across the entire domain. The present criterion allows a weaker form of control, provided the parts without uniform convergence are both short and subject to a common bound.

Takeaway: If a pointwise-convergent sequence of Riemann integrable functions has a Riemann integrable limit, a common bound, and uniform convergence outside intervals of arbitrarily small total length, then the integrals converge to the integral of the limit.

Check Your Understanding

Use the error estimates and examples in this tutorial to answer the following questions.

In the theorem, why must the total length of the exceptional intervals be small as well as the functions being uniformly bounded?
For the sequence $g_n$, what is the pointwise limit at $x=1/3$, and what is it at any other point?
Why does the sequence $f_n(x)=x^2/(x^2+1/n)$ converge uniformly on $[\delta,1]$ for each fixed $\delta>0$?
Which property of $h_n$ makes its integral zero for every $n$?
Does pointwise convergence by itself justify passing a limit through a Riemann integral? Explain briefly.

Interchanging Limits and Integrals

What You'll Learn