Differentiable everywhere with a bounded derivative — yet the derivative is not Riemann integrable. A sum of (x−a)2(b−x)2sinx−a1-type bumps over a fat Cantor set builds this legendary counterexample.
On a complementary interval (a,b), let φ(x)=(x−a)2(b−x)2sinx−a1 for a<x<b and φ=0 at the endpoints. Prove φ is differentiable everywhere with φ(a)=φ(b)=φ′(a)=φ′(b)=0: compute the derivative directly inside (a,b), and use the difference quotient at the endpoints. (Warning: φ′ is discontinuous at the left endpoint a — the proof reveals why.)
Stage N=4, zoomed on the first complementary interval (0.375,0.625): φ1 looks smooth, but the oscillatory term −(b−x)2cosx−a1 of its derivative keeps amplitude (b−a)2=0.0625 as x→0.375+, refusing to die.Show proof
Write u=x−a, v=b−x (so u+v=b−a and u,v>0); inside, φ=u2v2sinu1. The endpoint values follow by squeeze: ∣u2v2sinu1∣≤u2v2→0 as x→a+ or x→b−. Outside the interval φ≡0, so φ is continuous on [0,1] with φ(a)=φ(b)=0.
Inside, differentiate directly with the product and chain rules: dxd[u2v2]=2uv2−2u2v=2uv(v−u) (using dxdv=−1), and dxdsinu1=−u21cosu1, giving φ′(x)=2uv(v−u)sinu1−v2cosu1 (the display formula above).
At the endpoints use the difference quotient, not the limit of the formula above. For h>0, hφ(a+h)−φ(a)=h2(b−a−h)2sinh1, squeezed to 0 by h2(b−a)2, so φ′(a)=0. Likewise −hφ(b−h)−φ(b)=−h(b−h−a)2sinb−h−a1→0, so φ′(b)=0.
But φ′ is not continuous at a: at xm=a+2πm1 we have sinxm−a1=0 and cosxm−a1=1, so φ′(xm)=−(b−xm)2→−(b−a)2=0=φ′(a). The oscillatory term −(b−x)2cosx−a1 keeps amplitude (b−a)2 as x→a+ and never dies. (At the right endpoint b, the factor (b−x)2→0 kills the oscillation, so φ′ is continuous there.) Hence each bump is differentiable everywhere with value and derivative 0 at the endpoints, but not C1 — and exactly this failure is what makes f′ discontinuous on the Cantor set in Problem 3.
Let f(x)=∑k2−kφk(x), summed over all complementary intervals. Apply the Weierstrass M-test to ∑k2−kφk and ∑k2−kφk′ to prove f is differentiable at every point of [0,1] with bounded f′. The key is the k-uniform bounds ∣φk∣≤161 and ∣φk′∣≤23 (using (bk−ak)≤1).
Stage N=4: 15 complementary intervals, removed length 3215=0.46875, remaining length 3217=0.53125. On the sample grid sup∣f′∣≈0.0307 — comfortably inside the proved bound 3/2.Show proof
Each gk=2−kφk is differentiable on [0,1] by Problem 1 (endpoints included, derivative 0). With u,v≥0 and u+v=bk−ak≤1, AM-GM gives u2v2≤(2u+v)4≤161, so ∣gk∣≤2−k/16. Since ∑2−k/16<∞, the M-test makes ∑gk converge uniformly on [0,1] — the limit f is continuous.
The derivatives obey the same treatment: from ∣φk′∣≤2uv∣v−u∣+v2, using uv≤(2u+v)2≤41, ∣v−u∣≤u+v≤1, and v2≤1, we get ∣φk′(x)∣≤21+1=23 — uniform in k and x. Hence ∣gk′∣≤2−k⋅23, and the M-test makes ∑gk′=∑2−kφk′ converge uniformly as well.
Now the standard theorem: if a series of differentiable functions converges at one point and the series of derivatives converges uniformly, the sum is differentiable everywhere and its derivative is the sum of the derivatives. Hence f is differentiable at every point of [0,1] with f′(x)=∑k2−kφk′(x), bounded by ∣f′(x)∣≤∑k2−k⋅23=23. Numerically: sup∣f′∣≈0.0307 on the stage-4 grid.
Corollary: at any c in the Cantor set C, every φk and φk′ vanishes (as c lies outside all complementary intervals), so f(c)=0 and f′(c)=0. That is, f′ is identically 0 on C — yet, as Problem 3 shows, discontinuous at every point of C.
Prove f′ is discontinuous at every c in the Cantor set C: each neighborhood of c contains a whole complementary interval (ak,bk) (C has empty interior), on which f′=2−kφk′ oscillates near ak with amplitude ≈2−k(bk−ak)2, while f′(c)=0. Then conclude via Lebesgue's criterion (cited as a theorem below — proof at idea level; see the measure/lebesgue scenes) that f′ is not Riemann integrable.
(Lebesgue, cited): g bounded on [a,b]
g∈R[a,b]⟺g is continuous a.e.
xm=ak+πm1
f′(xm)=2−k(bk−xm)2(−1)m+1→±2−k(bk−ak)2
Stage N=4, zoomed on the first interval (0.375,0.625): as x→0.375+, the peak-to-peak swing of f′ stays ≈0.0625 while f′(c)=0 at every point of the Cantor set — so f′ is discontinuous at every point of C.Show proof
Lemma: every neighborhood of c∈C contains a whole complementary interval. Let In be the stage-n surviving closed interval containing c; then ∣In∣→0, and at stage n the middle chunk removed from In−1 is a complementary interval Jn=(ak,bk) of length 4−n. For n with ∣In−1∣<δ, all of Jn⊂In−1 lies within δ of c. So C has empty interior (every open interval eventually has its middle removed).
On such an interval (a,b)=(ak,bk), f′=2−kφk′ (every other term is 0). With xm=a+πm1, we have sinxm−a1=sin(πm)=0 and cosxm−a1=(−1)m, so φk′(xm)=−(b−xm)2(−1)m and f′(xm)=2−k(b−xm)2(−1)m+1→±2−k(b−a)2 (sign alternating with the parity of m).
But Problem 2's corollary gives f′(c)=0. Since xm→a and a is arbitrarily close to c, there are points arbitrarily close to c where f′ takes values both above +2−k(b−a)2/2 and below −2−k(b−a)2/2 — it cannot converge to f′(c)=0. Hence f′ is discontinuous at c, and c∈C was arbitrary: f′ is discontinuous at every point of C. Numerically: on the first interval (0.375,0.625), a dense grid near x→0.375+ shows maxf′−minf′≈0.0625=2⋅21⋅(0.25)2.
Finally we cite Lebesgue's criterion as a theorem (its proof is measure theory, so we keep it at idea level — see the measure/lebesgue scenes): a bounded function on [a,b] is Riemann integrable iff it is continuous almost everywhere, i.e. its discontinuity set has Lebesgue measure 0. Here f′ is bounded (Problem 2), its discontinuity set contains C, and C is closed with measure 1/2 (the complement's total length is limN→∞21−2−N=21). A discontinuity set of positive measure means f′ is not Riemann integrable. The punchline: a function can be differentiable everywhere with bounded derivative, yet its derivative need not be Riemann integrable.