Given [latex]f : [0, 1] → \mathbb{R}, f (x) = x,[/latex] evaluate [latex]\int_{[0,1]} {f} dm.[/latex]

The sequence of simple functions [latex]\varphi_{n} = \sum\limits_{i=1}^{2^{n}}{\frac{i − 1}{2^{n}} χ_{[(i−1)/2^{n},i/2^{n})}}[/latex] clearly lies in [latex]\mathcal{S}_{+}[/latex] and satisfies [latex]\varphi_{n} \nearrow f[/latex] (see Theorem 10.20 ). By Theorem 11.1 , [latex]\int_{[0,1]}{f} dm = \underset{n→∞}{\lim} \int_{[0,1]}{\varphi_{n}}dm[/latex] [latex]= \underset{n→∞}{\lim} \sum\limits_{i=1} ^{2^{n}}{\frac{i − 1}{2^{n}}} m \left(\left[\frac{i − 1}{2^{n}} , \frac{i}{2^{n}}\right)\right)[/latex] [latex]= \underset{n→∞}{\lim} \frac{1}{2^{2n}} \sum\limits_{i=1} ^{2^{n}}{(i − 1)}[/latex] [latex]= \underset{n→∞}{\lim} \frac{1}{2^{2n}} \frac{2^{n}(2^{n} − 1)}{2} = \frac{1}{2}.[/latex]

Question 11.1: Given f : [0, 1] → R, f(x) = x, evaluate ∫[0,1] f dm.

Elements of Real Analysis

Given $f : [0, 1] → \mathbb{R}, f (x) = x,$ evaluate $\int_{[0,1]} {f} dm.$

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

The sequence of simple functions

$\varphi_{n} = \sum\limits_{i=1}^{2^{n}}{\frac{i − 1}{2^{n}} χ_{[(i−1)/2^{n},i/2^{n})}}$

clearly lies in $\mathcal{S}_{+}$ and satisfies $\varphi_{n} \nearrow f$ (see Theorem 10.20). By Theorem 11.1,

$\int_{[0,1]}{f} dm = \underset{n→∞}{\lim} \int_{[0,1]}{\varphi_{n}}dm$

$= \underset{n→∞}{\lim} \sum\limits_{i=1} ^{2^{n}}{\frac{i − 1}{2^{n}}} m \left(\left[\frac{i − 1}{2^{n}} , \frac{i}{2^{n}}\right)\right)$

$= \underset{n→∞}{\lim} \frac{1}{2^{2n}} \sum\limits_{i=1} ^{2^{n}}{(i − 1)}$

$= \underset{n→∞}{\lim} \frac{1}{2^{2n}} \frac{2^{n}(2^{n} − 1)}{2} = \frac{1}{2}.$