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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

1.1A1choose

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 $x \in [a, b]$ .

1.2L1choose

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

2.1step 1.1L1algebra

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.

3.1step 1.1step 1.2step 2.1L2algebra

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 $L (f, P) \geq L (f_{j}, P) - η (b - a)$ .

4.1step 3.1L1algebra

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.

5.1step 4.1A1choose

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 $x \in [a, b]$ .

6.1step 4.1step 5.1L3algebra

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

7.1step 4.1step 6.1∎

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

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources