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A uniform limit of Riemann-integrable functions is Riemann integrable, and its integral is the limit of their integrals

Statement

Let $a < b$ be reals. Suppose every $f_{k} : [a, b] \to R$ is Riemann integrable and $f_{k} \to f$ uniformly on $[a, b]$ . Then $f$ is Riemann integrable and

$\int_{a}^{b} f_{k} ⟶ \int_{a}^{b} f .$

Facts & Assumptions

Given: Reals $a < b$ , integrable functions $f_{k} : [a, b] \to R$ , and uniform convergence $f_{k} \to f$ .

[A1]

Uniform convergence means that for every real $η > 0$ one index makes $∣ f_{k} (x) - f (x) ∣ < η$ for every later $k$ and every $x \in [a, b]$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[L1]

An integrable function on $[a, b]$ is bounded; conversely, a bounded function $h$ there is Riemann integrable exactly when, for every real $ε > 0$ , some partition $P$ satisfies $U (h, P) - L (h, P) < ε$ (The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ , Riemann's criterion: a bounded $f$ on $[a, b]$ is Darboux integrable if and only if for every real $ε > 0$ there is a partition $P$ with $U (f, P) - L (f, P) < ε$ ).

[L2]

Darboux upper and lower sums are finite sums of the subinterval suprema and infima times the subinterval lengths; finite sums preserve inequalities and split and telescope in the usual way (For bounded $f$ on $[a, b]$ and a partition $P$ : the infimum $m_{i}$ and supremum $M_{i}$ of $f$ on the $i$ -th subinterval, and the lower and upper Darboux sums $L (f, P) = \sum_{i} m_{i} Δ_{i}$ and $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , Laws of finite sums and finite products).

[L3]

If two integrable functions differ by at most $η$ uniformly, then their integrals differ by at most $η (b - a)$ (Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error).

Proof

technique · direct

1.1

Let $ε > 0$ be real, put $η : = ε / (4 (b - a)) > 0$ , and choose an index $j$ such that $∣ f_{j} (x) - f (x) ∣ < η$ for every $x \in [a, b]$ .

A1choose

1.2

By integrability of $f_{j}$ and [L1], choose a partition $P$ with $U (f_{j}, P) - L (f_{j}, P) < ε / 2$ .

L1choose

2.1

The integrable function $f_{j}$ is bounded, say $∣ f_{j} (x) ∣ \leq M$ on $[a, b]$ ; then $∣ f (x) ∣ \leq M + η$ , so $f$ is bounded.

step 1.1L1algebra

3.1

On each subinterval of $P$ , step 1.1 gives $\sup f \leq \sup f_{j} + η$ and $\inf f \geq \inf f_{j} - η$ ; these suprema and infima exist by step 2.1. Multiplying by the nonnegative subinterval lengths and summing gives $U (f, P) \leq U (f_{j}, P) + η (b - a)$ and $L (f, P) \geq L (f_{j}, P) - η (b - a)$ .

step 1.1step 1.2step 2.1L2algebra

4.1

Therefore $U (f, P) - L (f, P) \leq U (f_{j}, P) - L (f_{j}, P) + 2 η (b - a) < ε$ , so [L1] makes $f$ integrable.

step 3.1L1algebra

5.1

Now let $ε > 0$ be real and choose $N$ such that $∣ f_{k} (x) - f (x) ∣ < ε / (b - a + 1)$ for every $k \geq N$ and every $x \in [a, b]$ .

step 4.1A1choose

6.1

For $k \geq N$ , both functions are integrable, so [L3] gives $∣ \int_{a}^{b} f_{k} - \int_{a}^{b} f ∣ \leq ε (b - a) / (b - a + 1) < ε$ .

step 4.1step 5.1L3algebra

7.1

Step 6.1 proves $\int_{a}^{b} f_{k} \to \int_{a}^{b} f$ , while step 4.1 proves integrability of $f$ .

step 4.1step 6.1∎

Depends on

Used by

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

MIT OpenCourseWare 18.100B, Real Analysis, Lectures 20–21 (standard reference, not scraped)
W. Trench, Introduction to Real Analysis (standard reference, not scraped)