Suppose \(f , g : [ 0 , 1 ] \to \mathbb { R }\) with \(f\) Riemann integrable and
\[ | g ( x ) - g ( y ) | \leq | f ( x ) - f ( y ) | , \qquad x , y \in [ 0 , 1 ] . \]
Prove that \(g\) is also Riemann integrable.
Solution
1. The function \(g\) is bounded on \([0,1]\).
Since \(f\) is Riemann integrable, it is bounded. Thus there is \(M\geq0\) such that \[ \abs{f(x)}\leq M \] for every \(x\in[0,1]\).
Fix \(y_0\in[0,1]\). For every \(x\in[0,1]\), \[ \begin{aligned} \abs{g(x)} &\leq \abs{g(y_0)} + \abs{g(x)-g(y_0)}\\ &\leq \abs{g(y_0)} + \abs{f(x)-f(y_0)}\\ &\leq \abs{g(y_0)}+2M. \end{aligned} \] Hence \(g\) is bounded.
2. For every nonempty interval \(I\subseteq[0,1]\), the oscillations satisfy\[ \operatorname{osc}_I(g) \leq \operatorname{osc}_I(f), \]where\[ \operatorname{osc}_I(h) = \sup_I h-\inf_I h. \]
For all \(x,y\in I\), the hypothesis gives \[ \abs{g(x)-g(y)} \leq \abs{f(x)-f(y)} \leq \sup_I f-\inf_I f. \] Taking the supremum over \(x,y\in I\) on the left gives \[ \sup_I g-\inf_I g \leq \sup_I f-\inf_I f. \]
3. For every partition\[ P:\quad 0=x_0<x_1<\cdots<x_n=1, \]one has\[ U(g,P)-L(g,P) \leq U(f,P)-L(f,P). \]
Let \[ I_j=[x_{j-1},x_j]. \] By the definitions of the upper and lower Darboux sums, \[ U(h,P)-L(h,P) = \sum_{j=1}^n \operatorname{osc}_{I_j}(h)(x_j-x_{j-1}) \] for any bounded real-valued function \(h\). Applying step 2 to each \(I_j\) gives the displayed inequality.
4. For every \(\varepsilon>0\), there is a partition \(P\) such that\[ U(g,P)-L(g,P)<\varepsilon. \]
Since \(f\) is Riemann integrable, the Darboux criterion gives a partition \(P\) with \[ U(f,P)-L(f,P)<\varepsilon. \] Step 3 then gives \[ U(g,P)-L(g,P) \leq U(f,P)-L(f,P) < \varepsilon. \]
5. The function \(g\) is Riemann integrable on \([0,1]\).
Step 1 gives boundedness, and step 4 verifies the Darboux criterion for Riemann integrability.
6. Q.E.D.
Step 5 is the required conclusion.