Use the definition of the derivative to prove that if \(f\) and \(g\) are differentiable at \(x\), then \(fg\) is differentiable at \(x\).
Use the definition of the Riemann integral to prove that if \(f\) is bounded on \([a,b]\) and is continuous everywhere except for finitely many points in \((a,b)\), then \(f\in\mathscr{R}\) on \([a,b]\).
Solution
1. If \(f\) and \(g\) are differentiable at \(x\), then \(fg\) is differentiable at \(x\) with \((fg)'(x)=f'(x)g(x)+f(x)g'(x)\).
Assume that \(f\) and \(g\) are differentiable at \(x\). For \(h\ne0\), \[ \frac{f(x+h)g(x+h)-f(x)g(x)}{h} = f(x+h)\frac{g(x+h)-g(x)}{h} + g(x)\frac{f(x+h)-f(x)}{h}. \] Differentiability of \(f\) at \(x\) implies continuity there, so \[ f(x+h)\to f(x). \] Also, \[ \frac{g(x+h)-g(x)}{h}\to g'(x), \qquad \frac{f(x+h)-f(x)}{h}\to f'(x). \] Therefore \[ \lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} = f(x)g'(x)+g(x)f'(x). \] Thus \(fg\) is differentiable at \(x\), with \[ \boxed{(fg)'(x)=f'(x)g(x)+f(x)g'(x).} \]
For part (b), let \(f\) be bounded on \([a,b]\) and continuous except at finitely many points of \((a,b)\).
2. For every \(\varepsilon>0\) there are pairwise disjoint open intervals \(I_1,\dots,I_N\) around the discontinuities such that, for any partition \(P\) containing their endpoints, the partition intervals inside \(\bigcup_j I_j\) contribute less than \(\varepsilon/2\) to \(U(f,P)-L(f,P)\).
Let the discontinuity points of \(f\) in \((a,b)\) be \[ c_1,\dots,c_N. \] Because \(f\) is bounded, choose \(M>0\) such that \[ |f(x)|\le M \qquad(x\in[a,b]). \] Fix \(\varepsilon>0\).
Choose pairwise disjoint open intervals \(I_j\) around \(c_j\) whose total length is less than \[ \frac{\varepsilon}{4M} \] (if \(M=0\), the conclusion is immediate). The contribution to any upper-minus-lower Darboux sum from subintervals contained in these neighborhoods is at most \[ 2M\sum_{j=1}^N |I_j|<\frac\varepsilon2. \]
3. \(f\in\mathscr R\) on \([a,b]\).
Remove the chosen neighborhoods from \([a,b]\). The remaining set is a finite union of compact intervals on each of which \(f\) is continuous, hence uniformly continuous.
Therefore there exists \(\delta>0\) such that whenever \(x,y\) lie in the same remaining compact component and \(|x-y|<\delta\), one has \[ |f(x)-f(y)|<\frac{\varepsilon}{2(b-a)}. \] Choose a partition \(P\) containing all endpoints of the intervals \(I_j\) and subdividing every complementary compact interval into pieces of length less than \(\delta\).
On each complementary partition interval \(J\), the oscillation of \(f\) is less than \(\varepsilon/(2(b-a))\). Hence the total contribution of all such intervals to \[ U(f,P)-L(f,P) \] is at most \[ \frac{\varepsilon}{2(b-a)}(b-a)=\frac\varepsilon2. \] Combining this with step 2 gives \[ U(f,P)-L(f,P)<\varepsilon. \] Since \(\varepsilon>0\) was arbitrary, the Darboux criterion for Riemann integrability implies \[ \boxed{f\in\mathscr R([a,b]).} \]