Write a Proof Without Using the Textbook

Build the Proof From the Claim

Writing a proof without using a textbook does not mean refusing to use results established earlier in the course. It means making the logical route visible: identify the exact claim, choose appropriate established facts, and show how each fact applies to the objects in the problem. A reader should be able to follow the argument without needing an unspoken step supplied by a worked solution.

We will practise that approach on a question about integration. Suppose continuous functions $f_n$ approximate a continuous function $f$ uniformly on a closed interval. Does their integrals approximate the integral of $f$? The proof should not begin by guessing what the answer ought to be. First identify what must be shown: for each $\varepsilon>0$, the difference between the integrals must eventually be less than $\varepsilon$. Then find an estimate that connects that difference to the given uniform approximation.

The relevant bridge is an error estimate for integrals. The Absolute-Value Bound for an Integral, established earlier in the course, bounds the absolute value of an integral by the integral of the absolute value. The elementary order property of the Riemann integral then lets a pointwise bound control that latter integral. We prove the resulting estimate before applying it to a sequence.

State the target.
Translate convergence of integrals into an arbitrary-$\varepsilon$ requirement.

Find a quantitative bridge.
Relate the difference of integrals to a uniform bound on the difference of functions.

Use uniform convergence.
Choose an index that makes the function error small enough for the integral error target.

Check prerequisites.
Verify that every function being integrated is Riemann integrable.

A Quantitative Estimate for Integral Errors

The estimate below is useful well beyond sequences. It says that if two continuous functions are everywhere close, then their integrals over an interval are close, with the length of the interval determining the size of the possible accumulated error. Its proof is short, but each inequality has a role: the first uses a previously established theorem, and the second uses the pointwise hypothesis.

Theorem (Uniform Error Bound for Integrals): Let $a<b$, and let $f,g:[a,b]\to\mathbb R$ be continuous. Suppose $|f(x)-g(x)|\leq\eta$ for every $x\in[a,b]$, where $\eta\geq0$. Then $$ \left|\int_a^b f(x)\,dx-\int_a^b g(x)\,dx\right|\leq\eta(b-a). $$

Proof: The function $f-g$ is continuous, so it is Riemann integrable. By linearity of the integral and the Absolute-Value Bound for an Integral, $$ \left|\int_a^b f(x)\,dx-\int_a^b g(x)\,dx\right| =\left|\int_a^b (f(x)-g(x))\,dx\right| \leq\int_a^b |f(x)-g(x)|\,dx. $$ The hypothesis gives $0\leq |f(x)-g(x)|\leq\eta$ at every point. By the order property of the integral and the integral of a constant, $$ \int_a^b |f(x)-g(x)|\,dx\leq\int_a^b\eta\,dx=\eta(b-a). $$ Combining the inequalities proves the claim. $\square$

Notice the difference between a pointwise and a uniform estimate. A pointwise estimate at one selected $x$ would not bound the integral over the whole interval. Here the same $\eta$ works at every point, so the integral of the error is at most the constant error $\eta$ multiplied by the interval length.

Worked Example: Integrating a Uniformly Small Perturbation

On $[0,2]$, let $f(x)=x^2$ and $g(x)=x^2+x/20$. Their difference satisfies $$ |f(x)-g(x)|=\frac{x}{20}\leq\frac{2}{20}=\frac{1}{10} \qquad (0\leq x\leq2). $$ The Uniform Error Bound for Integrals, with $\eta=1/10$, gives $$ \left|\int_0^2 f(x)\,dx-\int_0^2 g(x)\,dx\right| \leq\frac{1}{10}(2-0)=\frac{1}{5}. $$ We can check the actual difference directly: $$ \int_0^2 g(x)\,dx-\int_0^2 f(x)\,dx =\int_0^2\frac{x}{20}\,dx =\left[\frac{x^2}{40}\right]_0^2 =\frac{4}{40} =\frac{1}{10}. $$ Thus the estimate is valid, though in this example it is not sharp.

Uniform Convergence Controls the Integrals

We can now prove the main result. Uniform convergence gives one index that controls the error at every point of the interval. The quantitative estimate turns that function error into an integral error. The only additional issue is existence of the limiting integral: each $f_n$ is continuous, and the Uniform Limits Preserve Continuity theorem established earlier shows that the uniform limit $f$ is continuous as well.

Theorem (Convergence of Integrals Under Uniform Convergence): Let $a<b$, and suppose $f_n:[a,b]\to\mathbb R$ is continuous for every positive integer $n$. If $f_n$ converges uniformly to $f:[a,b]\to\mathbb R$, then $f$ is continuous and $$ \lim_{n\to\infty}\int_a^b f_n(x)\,dx=\int_a^b f(x)\,dx. $$

Proof: Since each $f_n$ is continuous and the convergence is uniform, the Uniform Limits Preserve Continuity theorem implies that $f$ is continuous on $[a,b]$. Therefore $f$ and every $f_n$ are Riemann integrable there.

Let $\varepsilon>0$. Since $a<b$, the number $\varepsilon/(b-a)$ is positive. Uniform convergence gives a positive integer $N$ such that for every $n\geq N$ and every $x\in[a,b]$, $$ |f_n(x)-f(x)|<\frac{\varepsilon}{b-a}. $$ In particular, the non-strict bound required by the Uniform Error Bound for Integrals holds with $\eta=\varepsilon/(b-a)$. Applying that estimate gives, for every $n\geq N$, $$ \left|\int_a^b f_n(x)\,dx-\int_a^b f(x)\,dx\right| \leq\frac{\varepsilon}{b-a}(b-a)=\varepsilon. $$ To obtain the strict inequality in the definition of convergence, begin instead with $\varepsilon/2$ in place of $\varepsilon$. The same argument gives an index $N$ for which the absolute difference is at most $\varepsilon/2<\varepsilon$ whenever $n\geq N$. This proves convergence of the integrals. $\square$

The proof uses the interval length explicitly. On a longer interval, the same uniform error can contribute a larger integral error. This is why the tolerance for the function error is chosen as $\varepsilon/(b-a)$, rather than simply $\varepsilon$. It also illustrates a general proof-writing habit: when an estimate must be strictly less than a target, choosing half the target avoids a needless problem if an intermediate estimate gives only “less than or equal to.”

Worked Example: A Polynomial Sequence on a Closed Interval

Define $f_n(x)=x^2+x/n$ on $[0,2]$, and let $f(x)=x^2$. Each $f_n$ is continuous. For every $x\in[0,2]$, $$ |f_n(x)-f(x)|=\frac{x}{n}\leq\frac{2}{n}. $$ Since $2/n\to0$, this proves uniform convergence of $f_n$ to $f$. The theorem therefore gives convergence of the integrals. The quantitative estimate gives the more precise bound $$ \left|\int_0^2 f_n(x)\,dx-\int_0^2 f(x)\,dx\right| \leq\frac{2}{n}(2-0)=\frac{4}{n}. $$ Direct computation verifies the difference: $$ \int_0^2 f_n(x)\,dx-\int_0^2 f(x)\,dx =\frac{1}{n}\int_0^2x\,dx =\frac{1}{n}\left[\frac{x^2}{2}\right]_0^2 =\frac{2}{n}. $$ In particular, the difference tends to zero, as the theorem predicts.

Worked Example: A Trigonometric Perturbation

On $[0,\pi]$, define $g_n(x)=\cos x+1/n$, with $g(x)=\cos x$. Each function is continuous, and $$ |g_n(x)-g(x)|=\frac{1}{n} \qquad (0\leq x\leq\pi). $$ Thus $g_n$ converges uniformly to $g$. The integral estimate yields $$ \left|\int_0^\pi g_n(x)\,dx-\int_0^\pi g(x)\,dx\right| \leq\frac{\pi}{n}. $$ Indeed, since $\int_0^\pi\cos x\,dx=\sin(\pi)-\sin(0)=0$, $$ \int_0^\pi g_n(x)\,dx-\int_0^\pi g(x)\,dx =\int_0^\pi\frac{1}{n}\,dx =\frac{\pi}{n}. $$ The error bound is exact here: the perturbation has constant size $1/n$ across an interval of length $\pi$.

Worked Example: A Rational Approximation to a Polynomial

For $x\in[0,1]$, let $$ h_n(x)=\frac{x^3}{1+x^2/n}, \qquad h(x)=x^3. $$ The denominator $1+x^2/n$ is positive, so each $h_n$ is continuous. Their difference is $$ |h_n(x)-h(x)| =x^3\left|\,\frac{1}{1+x^2/n}-1\,\right| =\frac{x^5}{n+x^2}. $$ Because $0\leq x\leq1$, we have $x^5\leq1$ and $n+x^2\geq n$. Hence $$ |h_n(x)-h(x)|\leq\frac{1}{n} \qquad (0\leq x\leq1). $$ This proves uniform convergence. The integral estimate now gives $$ \left|\int_0^1 h_n(x)\,dx-\int_0^1 h(x)\,dx\right| \leq\frac{1}{n}(1-0)=\frac{1}{n}. $$ Since $\int_0^1h(x)\,dx=\int_0^1x^3\,dx=1/4$, the integral of $h_n$ differs from $1/4$ by at most $1/n$. No antiderivative for $h_n$ is needed.

What Makes the Argument Self-Contained?

A proof can rely on earlier theorems and still be complete. The important distinction is between citing a result that has already been established and skipping the step that explains why it applies. In the main proof, the continuity of $f$ is not assumed without justification: it follows from the Uniform Limits Preserve Continuity theorem. The integral estimate is proved before use, and the choice of $N$ is linked explicitly to the desired error.

The theorem is a sufficient condition, not a claim that uniform convergence is necessary for integrals to converge. It is also important that the functions here are continuous on the closed interval. That assumption ensures the integrals in the statement exist as Riemann integrals and allows us to use the earlier theorem about continuity of uniform limits. A different setting may require different integrability hypotheses and a different argument.

When writing a proof independently, check each step with questions like these: What is the exact conclusion to prove? Which hypothesis supplies each estimate? Is there a previously established result that can be cited by name? Does every quantity being integrated exist? Is the chosen tolerance strong enough to yield the required strict inequality? These checks keep a proof both concise and logically complete.

Check Your Understanding

For each question, identify the estimate or hypothesis that justifies the step.

Why does the proof of the Uniform Error Bound for Integrals use the Absolute-Value Bound for an Integral?
Where does the factor $b-a$ enter the estimate, and what does it represent?
In the convergence theorem, why must the limiting function $f$ be shown to be continuous before its Riemann integral is used?
Why is the function-error tolerance chosen in terms of $\varepsilon/(b-a)$?
For $h_n(x)=x^3/(1+x^2/n)$ on $[0,1]$, which inequalities give the uniform bound $1/n$?