Skip to content

3.26 Integrable function

  1. Show that the function \(f(x)=\frac{1}{1+x}\) is integrable on \([0,b]\) for all \(b>0\)​

    Proof

    1. One way:The function is continuous on a closed interval, then it is integrable by theorem

    2. Another way: Let \(P_{\varepsilon}=(0,\frac{b}{n},\ldots,n\frac{b}{n}=b)\), then \(U(f,P_{\varepsilon})=\sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)M_{i}=\sum_{i=1}^{n} \frac{b}{n}\frac{1}{1+x_{i}}=\frac{b}{n}\sum_{i=1}^{n}\frac{1}{1+\frac{\left(i-1\right)b}{n}} =\frac{b}{n}\sum_{i=1}^{n}\frac{n}{n+\left(i-1\right)b}\)

      And \(L(f,P_{\varepsilon})=\frac{b}{n}\sum_{i=1}^{n}\frac{n}{n+ib}\). Thus \(U(f,P_{\varepsilon})-L(f,P_{\varepsilon})=b\sum_{i=1}^{n}\left(\frac{1}{n+\left(i-1\right)b} -\frac{1}{n+ib}\right)=b\left(\frac{1}{n}-\frac{1}{n\left(1+b\right)}\right)=\frac{1}{n} \left(\frac{b^{2}}{1+b}\right)\)​

      Thus we choose \(n\geq N>\frac{1+b}{b^{2}}\varepsilon\)​

    Note that it's complex to prove integrable by this theorem, instead, we should use this epsilon theorem or the relation between continuity and integrability

  2. Let \(f:[a,b]\to \R\) be increasing 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 increasing on each \([x_{i-1}, x_i]\), so \(m_i = \inf \{ f(x) : x \in [x_{i-1}, x_i] \} = f(x_{i-1})\) and
    ​\(M_i = \sup \{ f(x) : x \in [x_{i-1}, x_i] \} = f(x_i)\)

    Then \(U(f,P) - L(f,P) = \sum_{i=1}^{n} (M_i - m_i)(x_i - x_{i-1})\) \(= \sum_{i=1}^{n} (x_i - x_{i-1}) (f(x_i) - f(x_{i-1}))\)

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

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

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

  3. Let \(f\) be a continuous non-negative function defined on \([a,b]\).

    Show that \(\displaystyle\lim_{n \to \infty}\left( \int_{a}^{b} f^{n}(x) \, dx \right)^{1/n}= \max\{ f(x) , x \in [a,b] \}\)

    Let \(x_{0} \in [a,b]\) such that \(f(x_{0}) = \max \{ f(x) : x \in [a,b] \} = M\)

    For all \(x \in [a,b]\), we have \(f(x) \leq M\), so \(\displaystyle \int_{a}^{b}f^{n}(x) \, dx \leq \int_{a}^{b}M^{n}\, dx = M^{n}(b-a )\)

    Thus, \(\displaystyle \left( \int_{a}^{b}f^{n}(x) \, dx \right)^{1/n}\leq (M(b-a))^{1/n}\). Thus \({\displaystyle\lim_{n\to\infty}\left(\int_{a}^{b}f^{n}(x)\,dx\right)^{1/n}\leq M}\)(1)

    Since \(f\) is continuous (in particular on \(x_{0}\)), we have that \(\forall\varepsilon > 0\), \(\exists\delta>0\) such that \(|x-x_{0}|<\delta:|f(x)-f(x_0)|<\varepsilon\)

    Then \(f(x) > M - \varepsilon\) when \(x\in[x_{0}-\delta,x_{0}+\delta]\). Thus, \(f^n(x) > (M - \varepsilon)^n\) for \(x\in[x_{0}-\delta,x_{0}+\delta]\)

    Thus \({{\displaystyle\int_{x_0-\delta}^{x_0+\delta}(M-\varepsilon)^{n}=2\delta(M-\varepsilon)^{n}<\int_{x_0-\delta}^{x_0+\delta}f^{n}(x)\,dx\leq\int_{a}^{b}f^{n}(x)\,dx}}\)

    Then \({{\displaystyle\lim_{n\to\infty}\left(2\delta\right)^{1/n}\left(M-\varepsilon\right)\leq\lim_{n\to\infty}\left(\int_{a}^{b}f^{n}(x)\,dx\right)^{1/n}}}\). Then \(\displaystyle M -\varepsilon < \lim_{n \to \infty}\left( \int_{a}^{b} f^{n}(x) \, dx \right)^{1/n}\)(2)

    From (1) and (2), we conclude \(\displaystyle M = \lim_{n \to \infty}\left( \int_{a}^{b} f^{n}(x) \, dx \right)^{1/n}\)

  4. Let \(f: [a,b] \to \mathbb{R}\) be integrable on \([a,b]\).
    Show that if one value of \(f(x)\) changes at some point \(x_0 \in [a,b]\), then \(f\) is still integrable and it integrates to the same value as before.

    Proof

    Since \(f\) is integrable, so \(\displaystyle U(f) = L(f) = \int_{a}^{b} f\).

    Let \(f_0\) be the modified function where we change the value of \(f\) at \(x_0 \in [a,b]\).

    Define \(d := |f(x_{0}) - f_{0}(x_{0})| \geq 0\).

    Let's prove that \(U(f) = U(f_{0})\) and \(L(f) =L(f_{0})\).

    Let \(\epsilon > 0\). To show that \(U(f) = U(f_0)\), it's enough to find a partition \(P\) such that \(U(f_{0},P)\leq U(f)+\epsilon\) (definition of infimum)

    \(U(f) = \inf \{ U(f,P) : P \in \mathcal{P} \}\), then there exists some partition \(P\) such that \(U(f,P) \leq U(f) + \frac{\epsilon}{2}\).

    Now, let \(P'\) be a refinement of \(P\) with the property that the interval(s) that contain \(x_0\) have length less than \(\frac{\epsilon}{4d}\).

    Because \(P \subseteq P'\), we have that \(U(f,P') \leq U(f,P)\).

    Since \(f\) and \(f_0\) agree on all points of \([a,b]\) except \(x_0\), \(|U(f,P') - U(f_0,P')| < 2d (x_i - x_{i-1}) < 2d \cdot \frac{\epsilon}{4d} = \frac{\epsilon}{2}\).

    (The "2" is in case \(x_0\) is in two subintervals of \(P'\).)

    Finally, we get that \(U(f_0, P') < U(f, P') + \frac{\epsilon}{2} < (U(f) + \frac{\epsilon}{2}) + \frac{\epsilon}{2} = U(f) + \epsilon\)

    Then \(U(f_0) = U(f)\).

    In a similar way, \(L(f) = L(f_0)\).

‍