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Homework 4

  1. (10 points)
    Show that the function \(f(x)\) defined as \(f(x) =\begin{cases} x, & \text{if } x \notin \mathbb{Q} \\ 0, & \text{if } x \in \mathbb{Q}\end{cases}\) is not integrable on \([0, 1]\).

    Let \(P\) be an arbitrary partition of \([0,1]\), \(0 = x_0 < x_1 < \dots < x_n = 1\).

    \(m_i(f, P) = 0\), \(M_{i}(f,P)=x_{i}\) (since \(\mathbb{Q}\) is dense in \(\mathbb{R}\))

    Then \(L(f, P) = \sum_{i=1}^{n} m_i (x_i - x_{i-1}) = 0\), \(U(f,P)=\sum_{i=1}^{n}x_{i}(x_{i}-x_{i-1})\) which is upper sum of \(f(x)=x\), then \(U(f)=\frac{1}{2}\)​

    Since \(P\) is an arbitrary partition of \([0,1]\), we have that \(L(f)=0\quad\text{and}\quad U(f)=\frac{1}{2}\)

    Thus, \(f\) is not integrable on \([0,1]\).

  2. (10 points)
    Let \(f : [a, b] \to \mathbb{R}\) be a decreasing function on \([a, b]\). Show that \(f\) is integrable on \([a, b]\).

    Let \(P\) be a partition of \([a,b]\) where all subintervals \([x_{i-1}, x_i]\) have the same length. We know that \(f\) is decreasing on each \([x_{i-1}, x_i]\), so \(m_{i}=\inf\{f(x):x\in[x_{i-1},x_{i}]\}=f(x_{i})\) and
    ​\(M_{i}=\sup\{f(x):x\in[x_{i-1},x_{i}]\}=f(x_{i-1})\)

    Then \(U(f,P) - L(f,P) = \sum_{i=1}^{n}(M_{i}- m_{i})(x_{i}- x_{i-1})\) \(=\sum_{i=1}^{n}(x_{i}-x_{i-1})(f(x_{i-1})-f(x_{i}))\)

    \(=(x_{i}-x_{i-1})(f(a)-f(b))\)

    Let \(\varepsilon > 0\). Choose a partition \(P_\varepsilon\) such that \(x_{i}-x_{i-1}<\frac{\varepsilon}{f(a)-f(b)},\forall i\)

    Then \(U(f, P_\varepsilon) - L(f, P_\varepsilon) < \varepsilon\). So, by theorem, \(f\) is integrable.

  3. (20 points)
    Let \(f(x) = \begin{cases} 1, & \text{if } x = \frac{1}{n} \text{ for some } n \in \mathbb{N} \\ 0, & \text{otherwise}. \end{cases}\)

    Show that \(f\) is integrable on \([0,1]\) and compute its integral.

    Solution

    Let's consider a partition \(P_{n}:\{0,\frac{1}{n},\frac{1}{n-1},\ldots,\frac{1}{2},1\}\)​

    Then for each subinterval \([\frac{1}{i},\frac{1}{i-1}]\), the length is \(\frac{1}{i-1}-\frac{1}{i}\), then split the length into \(n\) part, consider the partition \(\{\frac{1}{i},\frac{1}{i}+\frac{1}{n}\left(\frac{1}{i-1}-\frac{1}{i}\right),\frac{1}{i} +\frac{2}{n}\left(\frac{1}{i-1}-\frac{1}{i}\right),\ldots,\frac{1}{i}+\frac{n-1}{n} \left(\frac{1}{i-1}-\frac{1}{i}\right),\frac{1}{i-1}\}\)

    Then we do the same step for \(i=1,...,n\) and \(0\)​

    Then we have \(P_{n}^{\prime}\supset P_{n}\) and \(P_{\left({n+1}\right)^{n+1}}=\{0,\frac{1}{n^{2}},\frac{2}{n^{2}},\ldots,\frac{1}{n} ,\frac{1}{n}+\frac{1}{n\left(n-1\right)n},\ldots,\frac{1}{n}+\frac{n-1}{n\left(n-1\right)n} ,\frac{1}{n-1},\ldots,\frac{1}{2},\frac{1}{2}+\frac{1}{2n},\ldots,1\rbrace\)​

    By theorem we know \(U(f,P_{n})\leq U(f,P^{\prime}_{n})\) since \(P_{n}^{\prime}\supset P_{n}\)

    Then since \(m_{i}=\inf\{f(x):x\in[x_{i-1},x_{i}]\}=0\), then \(L(f,P_{n})=\sum_{i=1}^{n+1}m_{i}\left(x_{i}-x_{i-1}\right)=0\)

    Since \(M_{i}=\sup\{f(x):x\in[x_{i-1},x_{i}]\}\), then \(U(f,P_{n})\leq U\left(f,P_{n}^{\prime}\right)=\sum_{i=1}^{\left(n+1\right)^{n+1}} M_{i}\left(x_{i}-x_{i-1}\right)=1\left(\frac{1}{n\cdot n}\right)+1\cdot\left(\frac{1}{n\left(n-1\right)n} \right)+1\cdot\left(\frac{1}{n\left(n-1\right)n}\right)+\cdots+1\cdot\left(\frac{1}{\left(n-\left(n-1\right)\right)\left(n-\left(n-2\right)\right)n} \right)+1\cdot\left(\frac{1}{\left(n-\left(n-1\right)\right)\left(n-\left(n-2\right)\right)n} \right)\)

    \(=\frac{1}{n^{2}}+\frac{2}{n^{3}-n^{2}}+\cdots+\frac{1}{n}\)​

    And we know \(\lim_{n\to\infty}U(f,P_{n})\leq\lim_{n\to\infty}U\left(f,P_{n}^{\prime}\right)=0\) since \(U(f,P_{n})\leq U\left(f,P_{n}^{\prime}\right)\)​

    Thus \(\lim_{n\to\infty}U(f,P_{n})=\lim_{n\to\infty}L(f,P_{n})=0\)​

    Then \(f\) is integrable and \(\displaystyle \int^1_0 f(x)dx=0\)​

  4. (20 points)
    Prove that a bounded function \(f : [a, b] \to \mathbb{R}\) is Riemann integrable if and only if for every \(\epsilon > 0\), there exists a partition \(P\) of \([a, b]\) such that \(U(f, P) - L(f, P) < \epsilon.\)

    \(\Rightarrow\)) Assume that \(f\) is integrable. Then \(U(f) = L(f)\) where \(U(f) = \inf \{ U(f, P) \mid P \in \mathcal{P} \}\) and \(L(f) = \sup \{ L(f, P) \mid P \in \mathcal{P}\}\)

    Given \(\epsilon > 0\), there exist partitions: \(P_1 \in \mathcal{P}\) such that \(U(f, P_1) \leq U(f) + \frac{\epsilon}{2}\)

    \(P_2 \in \mathcal{P}\) such that \(L(f) - \frac{\epsilon}{2} \leq L(f, P_2)\) by definition of inf and supimage​

    Let \(P=P_{1}\cup P_{2}\Rightarrow\) \(L(f,P_{2})\leq L(f,P)\), \(U(f,P)\leq U(f,P_{1})\)

    \(\Rightarrow U(f,P)-L(f,P)\leq U(f,P_{1})-L(f,P)\) \(\leq U(f, P_1) - L(f, P_2) \leq U(f) + \frac{\varepsilon}{2} - L(f, P_2)\) \(\leq U(f) + \frac{\varepsilon}{2} - (L(f) - \frac{\varepsilon}{2}) = U(f) + \frac{\varepsilon}{2} - L(f) + \frac{\varepsilon}{2} = \varepsilon\)

    \(\Leftarrow\)) We want to show that \(U(f)=L(f)\)

    We know \(U(f)\leq U(f,P)\), \(L(f)\geq L(f,P)\)

    Hence \(0 \leq U(f) - L(f)\) \(\leq U(f,P)-L(f)\) \(\leq U(f,P)-L(f,P)<\varepsilon\)

    \(\Rightarrow U(f) - L(f) = 0\)

  5. (20 points)
    Prove that the function \(f(x) = \begin{cases} \frac{1}{q}, & \text{if } x = \frac{p}{q} \text{ for } p \in \mathbb{Z}, q \in \mathbb{N}, \text{ and } p, q \text{ are mutually prime} \\ 0, & \text{if } x \text{ is irrational}. \end{cases}\) is integrable on \([0,1]\).

    Consider \(P_{n}=\{0,\frac{1}{n},\frac{2}{n},\ldots,1\}\), then \(m_{i}=\inf\{f(x):x\in[x_{i-1},x_{i}]\}=0\) since irrational number is dense in any compact set.

    Then \(M_{i}=\sup\{f(x):x\in[x_{i-1},x_{i}]\}\) and \(U(f,P_{n})-L\left(f,P_{n}\right)=\sum_{i=1}^{n}\left(M_{i}-m_{i}\right)\left(x_{i} -x_{i-1}\right)=\sum_{i=1}^{n}M_{i}\left(x_{i}-x_{i-1}\right)\)

    Let's \(B_{N}=\{1,\frac{1}{2},\frac{1}{3},\frac{2}{3},...,\frac{N-1}{N}\}\) with \(|B_N|=m\), then consider \(P_{n}=\{x_0=0,x_1,...,x_{n-1},x_n=1\}\)

    We know \(B_{N}\subset [0,1]\), then \(\sum_{i=1}^{n}M_{i}\left(x_{i}-x_{i-1}\right)=\sum_{\left\lbrack x_{i-1},x_{i}\right\rbrack\cap B_{N}\neq\emptyset}M_{i}\left(x_{i}-x_{i-1}\right)+\sum_{\left\lbrack x_{i-1},x_{i}\right\rbrack\cap B_{N}=\emptyset}M_{i}^{\prime}\left(x_{i}-x_{i-1}\right)\)​

    Since \(M_{i}=\sup\{f(x):x\in[x_{i-1},x_{i}]\cap B_{N}\}=\frac{1}{i-1}\) and \(M_{i}^{\prime}=\sup\{f(x):x\in[x_{i-1},x_{i}]\setminus B_{N}\}\leq\frac{1}{N+1}\)

    And we know there are at most \(2m\) subintervals such that \(\left\lbrack x_{i-1},x_{i}\right\rbrack\cap B_{N}\neq\emptyset\) because one point in \(B_N\) can be in two intervals

    Then \(\sum_{i=1}^{n}M_{i}\left(x_{i}-x_{i-1}\right)\leq2m\cdot1\cdot\max\left(x_{i}-x_{i-1} \right)+\frac{1}{N+1}\sum\left(x_{i}-x_{i-1}\right)\leq2m\cdot1\cdot\max\left(x_{i} -x_{i-1}\right)+\frac{1}{N+1}\left(1-0\right)\leq2m\cdot1\cdot\max\left(x_{i}-x_{i-1} \right)+\frac{1}{N+1}\)

    Then let \(\max\left(x_{i}-x_{i-1}\right)<\frac{\varepsilon}{4m}\) and \(\frac{1}{N+1}<\frac{\varepsilon}{2}\), then we have \(\sum_{i=1}^{n}M_{i}\left(x_{i}-x_{i-1}\right)<\varepsilon\)​

    Thus \(U(f,P_{n})-L\left(f,P_{n}\right)<\varepsilon\), then \(f\) is integrable

  6. (20 points)
    Let \(f\) and \(g\) be two integrable functions on the closed interval \([a, b]\). Show that:

    1. The function \(f + g\) is integrable on \([a, b]\) and \(\displaystyle\int_{a}^{b} (f + g) \,dx = \int_{a}^{b} f(x) \,dx + \int_{a}^{b} g(x) \,dx.\)

      Let \(P=(x_0,...,x_n)\) be partition on \([a,b]\), then \(L(f,P)=\sum_{i=0}^{n}\left(x_{i}-x_{i-1}\right)m_{i}\) and \(U(f,P)=\sum_{i=0}^{n}\left(x_{i}-x_{i-1}\right)M_{i}\)

      Also, \(m_{i}=\inf\{f(x)\mid x\in[x_{i-1},x_{i}]\}\), \(m_{i}^{\prime}=\inf\{g(x)\mid x\in[x_{i-1},x_{i}]\}\)

      Then \(m_{i}+m_{i}^{\prime}=\inf\{f(x)\mid x\in[x_{i-1},x_{i}]\}+\inf\{g(x)\mid x\in[x_{i-1} ,x_{i}]\}\leq\inf\{\left(f+g)(x\right)\mid x\in[x_{i-1},x_{i}]\}=m_{i}^{*}\)

      Then \(L(f+g,P)=\sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)m_{i}^{*}\geq\sum_{i=1}^{n}\left (x_{i}-x_{i-1}\right)\left(m_{i}+m_{i}^{\prime}\right)=L(f,P)+L(g,P)\)

      Similarly, \(U(f+g,P)\leq U(f,P)+U(g,P)\)

      Then \(U(f+g,P_{n})-L(f+g,P_{n})\leq U(f,P_{n})+U(g,P_{n})-L(f,P_{n})-L\left(g,P_{n}\right )=U\left(f,P_{n}\right)-L\left(f,P_{n}\right)+U\left(g,P_{n}\right)-L\left(g,P_{n} \right)\)

      Since \(f,g\) is integrable, then \(U(f,P_{n})-L(f,P_{n})<\frac{\varepsilon}{2},U(g,P_{n})-L(g,P_{n})<\frac{\varepsilon}{2}\)​

      Thus \(U(f+g,P_{n})-L(f+g,P_{n})<\varepsilon\). Thus \(f+g\) is integrable.

      And \({\displaystyle\int_{a}^{b}(f+g)(x)dx=\lim_{n\to\infty}U(f+g,P_{n})}\) \(\displaystyle=\lim_{n\to\infty}U(f,P_{n})+\lim_{n\to\infty}U(g,P_{n})=\int_{a}^{b}f(x)dx +\int_{a}^{b}g(x)dx\) 2. The function \(|f|\) is integrable on \([a, b]\).

      Let \(P=(x_0,...,x_n)\) be partition on \([a,b]\), then \(L(\left|f\right|,P)=\sum_{i=0}^{n}\left(x_{i}-x_{i-1}\right)m_{i}\) and \(U(|f|,P)=\sum_{i=0}^{n}\left(x_{i}-x_{i-1}\right)M_{i}\)

      Also, \(m_{i}=\inf\{\left|f(x)\right|\mid x\in[x_{i-1},x_{i}]\}\) and \(M_{i}=\sup\{\left|f(x)\right|\mid x\in[x_{i-1},x_{i}]\}\)

      \(m'_{i}=\inf\{ f(x) \mid x\in[x_{i-1},x_{i}]\}\) and \(M_{i}=\sup\{ f(x) \mid x\in[x_{i-1},x_{i}]\}\)

      Then since \(|f(x)-f(y)|\geq ||f(x)|-|f(y)||\), then \(M_i-m_i\leq M_i'-m_i'\)

      Then \(U(\left|f\right|,P)-L(\left|f\right|,P)=\sum_{i=0}^{n}\left(x_{i}-x_{i-1}\right) \left(M_{i}-m_{i})\leq U(f,P\right)-L(f,P)\)

      Since \(f\) is integrable, then \(U(f,P)-L(f,P)<\varepsilon\)

      Thus \(U(\left|f\right|,P_{n})-L(\left|f\right|,P_{n})<\varepsilon\). Thus \(|f|\) is integrable.