Week 3 Tutorial Solutions

Integrating

1. Since $\mathbb{Q}$ is a Borel set the function $1_{\mathbb{Q}}$ is $(\mathsf{Bor}(\mathbb{R}),\mathsf{Bor}(\mathbb{R}))$ measurable. As $\mathbb{Q}$ is countable its Lebesgue measure is zero so the integral of $1_{\mathbb{Q}}$ is zero.
2. The function is continuous and therefore $(\mathsf{Bor}(\mathbb{R}),\mathsf{Bor}(\mathbb{R}))$ measurable. From properties of the sine function we get $$|\sin (n\pi +\frac{\pi }{2}+x)|\ge \frac{1}{2}$$ for all $n\in \mathbb{N}$ and and all $-\frac{1}{100}\le x\le \frac{1}{100}$ therefore $${\displaystyle \frac{|\sin (x)|}{x}}\ge {\displaystyle \frac{1}{4n\pi }}{1}_{[n\pi +\frac{\pi }{2}-\frac{1}{100},n\pi +\frac{\pi }{2}+\frac{1}{100}]}(x)$$ for all $n\in \mathbb{N}$. But then $$\int f\mathsf{d}\lambda \ge \sum _{n=1}^N{\displaystyle \frac{2}{100}}\cdot {\displaystyle \frac{1}{4n\pi }}$$ for every $N\in \mathbb{N}$ from which we conclude that the integral of $f$ is infinity.
3. The function is a product of a measurable simple function and a continuous function. It is therefore measurable. The function is bounded above by $1_{[0,1]}$ so has finite integral. For every $N\in \mathbb{N}$ we have the simple function $${\varphi }_N=\sum _{n=0}^{2^N-1}{1}_{[\frac{n}{2^N},\frac{n+1}{2^N}]}\cdot {\left({\displaystyle \frac{n}{2^N}}\right)}^2$$ which satisfies $0\le {\varphi }_N\le {\varphi }_{N+1}\le f$ for all $N\in \mathbb{N}$ and ${\varphi }_N\to f$ as $N\to \mathrm{\infty }$. Thus $$\int f\mathsf{d}\lambda =\underset{N\to \mathrm{\infty }}{lim}\int {\varphi }_N\mathsf{d}\lambda =\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{2^N}}\sum _{n=0}^{2^N-1}{\displaystyle \frac{n^2}{2^{2N}}}$$ the last of which is a limit of Riemann sums for the Riemann integral of $g(x)=x^2$ on $[0,1]$. Therefore, by the fundamental theorem of calculus, the answer is 1/3.
4. The function is a product of a measurable simple function and a function that is continuous on $(0,\mathrm{\infty })$. It is therefore measurable. Put $$f_n=1_{[\frac{1}{n},1-\frac{1}{n}]}\cdot f$$ and argue, just as in the previous solution, that $$\int f_n\mathsf{d}\lambda =\sqrt{1-{\displaystyle \frac{1}{n}}}-\sqrt{{\displaystyle \frac{1}{n}}}$$ can be calculated by evaluating the Riemann integral of $f_n$ on $[\frac{1}{n},1-\frac{1}{n}]$. We can then calculate $$\int f\mathsf{d}\lambda =\underset{n\to \mathrm{\infty }}{lim}\int f_n\mathsf{d}\lambda =1$$ using the monotone convergence theorem.
5. The function takes only countably many values, and takes each value on either a countable set or on $\mathbb{R}\setminus \mathbb{Q}$ so is therefore measurable. Its integral is zero because it is bounded above by the function $1_{\mathbb{Q}}$.
6. One can show by induction that each $f_n$ is continuous on $[0,1]$. As a pointwise limit of measurable functions we know from a worksheet that $g$ is measurable. Writing $I(n,1),\dots ,I(n,2^{n-1})$ for the open intervals removed at the $n$th stage of the construction of the middle-thirds Cantor set, we see that $$g\ge \sum _{n=1}^N\sum _{i=1}^{2^{n-1}}{\displaystyle \frac{2i-1}{2^n}}\cdot 1_{I(n,i)}$$ for all $N\in \mathbb{N}$ because the $I(n,i)$ air pairwise disjoint. Since each $I(n,i)$ has length $3^{-n}$ we conclude that $$\int g\mathsf{d}\lambda \ge \sum _{n=1}^N{\displaystyle \frac{1}{3^n}}\sum _{i=1}^{2^{n-1}}{\displaystyle \frac{2i-1}{2^n}}=\sum _{n=1}^N{\displaystyle \frac{1}{3^n}}$$ for all $N\in \mathbb{N}$. The integral of $g$ is therefore at least $\frac{1}{2}$. Since $g(1-x)=1-g(x)$ on $[0,1]$ the integral must be exactly $\frac{1}{2}$.
   
   Alternatively, note straight away that $g(1-x)=1-g(x)$ on $[0,1]$ and integrate this equation to conclude immediately that the integral is $\frac{1}{2}$.

The function $1_{\mathbb{Q}}$ is not Riemann integrable on any interval $[a,b]$ with $a<b$. Every other function is Riemann integrable on every closed interval, except that the function in question 4 is not Riemann integrable on any closed interval that contains zero.