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    Riemann integrability under $|g(x)-g(y)|\le|f(x)-f(y)|$

    P-IZZ3T

    Area
    Complex Analysis
    Topics
    Integrals, Riemann Integrability
    Status
    draft

    Suppose \(f, g: [0, 1] \to \RR\) where \(f\) is Riemann integrable and for \(x, y\in [0, 1]\), \[ \abs{g(x) - g(y)} \leq \abs{f(x) - f(y)} .\]

    Prove that \(g\) is Riemann integrable.

    Solution

    Because \(f\) is Riemann integrable on \([0,1]\), it is bounded. Fix \(x_0\in[0,1]\). Then \[ |g(x)|\le |g(x_0)|+|g(x)-g(x_0)| \le |g(x_0)|+|f(x)-f(x_0)|, \] so \(g\) is also bounded.

    For an interval \(I\subset[0,1]\), let \[ \omega_f(I)=\sup_{x,y\in I}|f(x)-f(y)|, \qquad \omega_g(I)=\sup_{x,y\in I}|g(x)-g(y)|. \] The hypothesis gives \[ \omega_g(I)\le \omega_f(I) \] for every \(I\).

    Since \(f\) is Riemann integrable, for every \(\varepsilon>0\) there is a partition \(P\) of \([0,1]\) such that \[ U(f,P)-L(f,P) =\sum_{I\in P}\omega_f(I)|I|<\varepsilon. \] For the same partition, \[ U(g,P)-L(g,P) =\sum_{I\in P}\omega_g(I)|I| \le\sum_{I\in P}\omega_f(I)|I| <\varepsilon. \] Hence the Darboux criterion shows that \(g\) is Riemann integrable.

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