Week 3 Tutorial Solutions

Integrating

1. Since $\Q$ is a Borel set the function $1_\Q$ is $(\borel(\R),\borel(\R))$ measurable. As $\Q$ is countable its Lebesgue measure is zero so the integral of $1_\Q$ is zero.
2. The function is continuous and therefore $(\borel(\R),\borel(\R))$ measurable. From properties of the sine function we get \[ |\sin(n\pi + \tfrac{\pi}{2} + x)| \ge \tfrac{1}{2} \] for all $n \in \N$ and and all $-\tfrac{1}{100} \le x \le \tfrac{1}{100}$ therefore \[ \dfrac{|\sin(x)|}{x} \ge \dfrac{1}{4 n \pi} 1_{\left[ n \pi + \tfrac{\pi}{2} - \tfrac{1}{100}, n \pi + \tfrac{\pi}{2} + \tfrac{1}{100} \right]}(x) \] for all $n \in \N$. But then \[ \int f \intd \lambda \ge \sum_{n=1}^N \dfrac{2}{100} \cdot \dfrac{1}{4n\pi} \] for every $N \in \N$ 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 $N \in \N$ we have the simple function \[ \phi_N = \sum_{n=0}^{2^N-1} 1_{\left[ \tfrac{n}{2^N}, \tfrac{n+1}{2^N} \right]} \cdot \left( \dfrac{n}{2^N} \right)^2 \] which satisfies $0 \le \phi_N \le \phi_{N+1} \le f$ for all $N \in \N$ and $\phi_N \to f$ as $N \to \infty$. Thus \[ \int f \intd \lambda = \lim_{N \to \infty} \int \phi_N \intd \lambda = \lim_{N \to \infty} \dfrac{1}{2^N} \sum_{n=0}^{2^N-1} \dfrac{n^2}{2^{2N}} \] 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 $(0,\infty)$. It is therefore measurable. Put \[ f_n = 1_{\left[ \tfrac{1}{n}, 1-\tfrac{1}{n} \right]} \cdot f \] and argue, just as in the previous solution, that \[ \int f_n \intd \lambda = \sqrt{1 - \dfrac{1}{n}} - \sqrt{\dfrac{1}{n}} \] can be calculated by evaluating the Riemann integral of $f_n$ on $[\tfrac{1}{n}, 1 - \tfrac{1}{n}]$. We can then calculate \[ \int f \intd \lambda = \lim_{n \to \infty} \int f_n \intd \lambda = 1 \] using the monotone convergence theorem.
5. The function takes only countably many values, and takes each value on either a countable set or on $\R \setminus \Q$ so is therefore measurable. Its integral is zero because it is bounded above by the function $1_\Q$.
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 \[ g \ge \sum_{n=1}^N \sum_{i=1}^{2^{n-1}} \dfrac{2i-1}{2^n} \cdot 1_{I(n,i)} \] for all $N \in \N$ because the $I(n,i)$ air pairwise disjoint. Since each $I(n,i)$ has length $3^{-n}$ we conclude that \[ \int g \intd \lambda \ge \sum_{n=1}^N \dfrac{1}{3^n} \sum_{i=1}^{2^{n-1}} \dfrac{2i - 1}{2^n} = \sum_{n=1}^N \dfrac{1}{3^n} \] for all $N \in \N$. The integral of $g$ is therefore at least $\tfrac{1}{2}$. Since $g(1-x) = 1-g(x)$ on $[0,1]$ the integral must be exactly $\tfrac{1}{2}$.
   
   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 $\tfrac{1}{2}$.

The function $1_\Q$ is not Riemann integrable on any interval $[a,b]$ with $a \l b$. 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.