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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 $f_k:[a,b]\to\mathbb{R}$ is Riemann integrable and $f_k\to f$ uniformly on $[a,b]$ . Then $f$ is Riemann integrable and

$\int_a^b f_k\longrightarrow\int_a^b f.$

Facts & Assumptions

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

[A1]

Uniform convergence means that for every real $\eta>0$ one index makes $|f_k(x)-f(x)|<\eta$ 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 $\varepsilon>0$ , some partition $P$ satisfies $U(h,P)-L(h,P)<\varepsilon$ (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 $\varepsilon > 0$ there is a partition $P$ with $U(f,P) - L(f,P) < \varepsilon$ ).

[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 $L(f,P) = \sum_i m_i \Delta_i$ and $U(f,P) = \sum_i M_i \Delta_i$ , Laws of finite sums and finite products).

[L3]

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

Proof

technique · direct

1.1

Let $\varepsilon>0$ be real, put $\eta:=\varepsilon/(4(b-a))>0$ , and choose an index $j$ such that $|f_j(x)-f(x)|<\eta$ for every $x\in[a,b]$ .

A1choose

1.2

By integrability of $f_j$ and [L1], choose a partition $P$ with $U(f_j,P)-L(f_j,P)<\varepsilon/2$ .

L1choose

2.1

The integrable function $f_j$ is bounded, say $|f_j(x)|\le M$ on $[a,b]$ ; then $|f(x)|\le M+\eta$ , so $f$ is bounded.

step 1.1L1algebra

3.1

On each subinterval of $P$ , step 1.1 gives $\sup f\le\sup f_j+\eta$ and $\inf f\ge\inf f_j-\eta$ ; these suprema and infima exist by step 2.1. Multiplying by the nonnegative subinterval lengths and summing gives $U(f,P)\le U(f_j,P)+\eta(b-a)$ and $L(f,P)\ge L(f_j,P)-\eta(b-a)$ .

step 1.1step 1.2step 2.1L2algebra

4.1

Therefore $U(f,P)-L(f,P)\le U(f_j,P)-L(f_j,P)+2\eta(b-a)<\varepsilon$ , so [L1] makes $f$ integrable.

step 3.1L1algebra

5.1

Now let $\varepsilon>0$ be real and choose $N$ such that $|f_k(x)-f(x)|<\varepsilon/(b-a+1)$ for every $k\ge N$ and every $x\in[a,b]$ .

step 4.1A1choose

6.1

For $k\ge N$ , both functions are integrable, so [L3] gives $\left|\int_a^b f_k-\int_a^b f\right|\le \varepsilon(b-a)/(b-a+1)<\varepsilon$ .

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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 67 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)