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Every null subset of the circle lies in the divergence set of the Fourier series of some continuous complex-valued function. One can arrange unbounded Fourier partial sums there. The compact-set Fourier amplification lemma, compact batching of a small open set and frequency-separated Fourier block series construct the function by a summable series of small blocks. The result covers nonclosed and dense null sets; it does not require the divergence set to equal the originally specified set.

Write $S_{N}f(t)={\sum }_{|n|\le N}\hat{f}(n)e^{int}$ for the symmetric Fourier partial sum, with normalized Fourier coefficients on $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$. We will prove the stronger conclusionIf $E$ is empty, take $f=0$. Otherwise the proof of the Kahane-Katznelson divergence theorem has three steps: a bounded trigonometric polynomial whose partial sum is large on short arcs, a finite-arc batching of an arbitrary null set, and a uniformly convergent sequence of separated frequency blocks. In particular, no assumption that $E$ is closed or compact is needed.

First consider a nonempty finite family of arcs with centers ${\theta }_{\nu }$, half-lengths $0<d_{\nu }\le 1$, and $D={\sum }_{\nu }{d}_{\nu }$. On the closed unit disk defineEach summand has positive real part there, and $\Phi (0)=1$. Thus the principal complex logarithm $A=\log \Phi$ is holomorphic on a neighborhood of the closed disk, $A(0)=0$, and $|\mathrm{Im}A|<\pi /2$. To bound its real part on an arc, put $h=t-{\theta }_{\nu }$ with $|h|\le d_{\nu }$ and $q=1+d_{\nu }-\cos h$. ThenAfter multiplication by the weight $d_{\nu }/D$, this contribution is at least $1/(4D)$; all the other contributions have positive real part. Throughout the arc union we therefore haveThis is the positive-real logarithmic amplifier for short arcs. Since $A$ is holomorphic beyond the closed unit disk, its Taylor series converges uniformly there. Choose a polynomial $Q(z)={\sum }_{\nu =1}^d{a}_{\nu }{z}^{\nu }$ with ${\sup }_{|z|\le 1}|Q(z)-A(z)|<1$. There is no constant term because $A(0)=0$. Set $C=\pi /2+1$ andwhere the integer $M>d$ may be chosen arbitrarily large. We have $\Vert P{\Vert }_{\infty }\le 1$, and its frequencies lie in $[M-d,M+d]$. At the cut $N=M$, the positive-frequency half of $Q$ is excluded and the negative-frequency half is included. Henceon the arc union. In particular, for any $K>0$, a total half-length $D\le \frac{1}{4}{e}^{-1-2CK}$ gives $|S_{M}P|\ge K$ there. This proves the needed compact-set Fourier amplification lemma, with the additional freedom to place its entire frequency block above any prescribed frequency.

Next we construct the null-set limsup cover by finite arc families. SetSince $E$ has Lebesgue measure zero, for every row $\ell \ge 1$ choose a countable arc cover $\{I_{\ell ,m}:m\ge 1\}$ of $E$ with positive half-lengths $d_{\ell ,m}$ and ${\sum }_m{d}_{\ell ,m}<2^{-\ell }{D}_1$. We can take closed arcs by slightly enlarging an open cover; their half-lengths are less than one. The whole two-index array has total half-length less than $D_1$. Absolute convergence of this sum lets us choose strictly increasing integers $N_j$ such that the sum outside the finite square $1\le \ell ,m\le N_j$ is less than $D_{j+1}$. Put $N_0=0$ and let ${\mathcal{F}}_j$ contain those arcs for whichEach family is finite. Its total half-length is less than $D_j$: for $j=1$ use the total sum, and for $j>1$ use the tail outside the preceding square. Moreover, every $t\in E$ belongs to arcs with unbounded row index, so belongs to the unions of infinitely many of the families ${\mathcal{F}}_j$. This batching is essential: a dense null set need not have a finite cover of small total length, whereas the infinitely-often finite covers above suffice.

Apply the amplification construction to ${\mathcal{F}}_j$ with height $K_j$, obtaining $P_j$ with $\Vert P_j{\Vert }_{\infty }\le 1$ and $|S_{M_j}{P}_j(t)|\ge K_j$ on that family's union. Choose the modulation integers successively so that the frequency intervals $[M_j-d_j,M_j+d_j]$ are positive and each lies strictly above the preceding one. Empty families may be represented by $P_j=0$ and an unused frequency interval. Define the frequency-separated Fourier block seriesThe bound ${\sum }_j{a}_j\Vert P_j{\Vert }_{\infty }\le 1$ gives absolute uniform convergence, so $f$ is continuous. Uniform convergence also justifies computing every Fourier coefficient term by term. At frequency $M_j$, all earlier blocks have been included in full and all later blocks are absent:For $t$ in the union of ${\mathcal{F}}_j$, the reverse triangle inequality givesEvery $t\in E$ lies in infinitely many such unions, and $M_j\to \infty$. Therefore the boxed unboundedness holds at every point of $E$, completing the proof for arbitrary null sets.

We give a quantitative construction for the whole open set, including the case of infinitely many component intervals. Pointwise divergence at a single point is insufficient here. Write $z=e^{2\pi it}$ and use normalized measure on $\mathbb{T}$.

First let $K$ be a finite union of closed arcs with $|K|\le 3\epsilon$, where $\epsilon$ is small. Choose a smooth function nonnegative function $u$, equal to one on $K$, with $u\le 1$ and $\eta :=\int u\le 6\epsilon$. This follows by making the transition regions around the finitely many arc endpoints sufficiently short. Define the analytic Herglotz integralIts real part is $(\eta +P[u](z))/(2\eta )>0$, where $P[u]$ is the disk Poisson integral, and $G(0)=1$. Smoothness of $u$ makes $G$ continuous and smooth function up to the circle; on $K$ its boundary real part is at least $1/(2\eta )$. The branch $H=\log G$ therefore has $H(0)=0$ andThe complex logarithm has no branch problem because $G$ lies in the right half-plane. Its smooth function boundary values give absolutely summable Taylor coefficients. Choose an analytic polynomial $A$, with zero constant coefficient, uniformly within $1/4$ of $H$ on the closed disk. Put $P(t)=\mathrm{Im}A(e^{2\pi it})/\pi$. Then $\Vert P{\Vert }_{\infty }\le 1/2+1/(4\pi )<1$.

If $D=\mathrm{deg}A$ and $m>D$, the modulated trigonometric polynomial $Q(t)=e^{2\pi imt}P(t)$ has spectrum in $[m-D,m+D]$. A sharp Fourier partial sum at $m$ selects exactly the negative-frequency half of $P$, since $A(0)=0$:Consequently, for every $t\in K$,The carrier frequency $m$ can be arbitrarily large. This compact-set Fourier amplification lemma uses a bounded imaginary part to obtain a large one-sided Fourier multiplier; no assertion about pointwise divergence is used.

Now let $E$ be any open set of measure $\delta$. DefineIf this is zero take $f=0$. Otherwise choose a decreasing sequence ${\delta }_j\downarrow 0$, with ${\delta }_1=\delta$, so small for $j\ge 2$ thatFor $j=1$ this inequality also holds: $B(\delta )>4$ and $B(\delta )/2\ge B(\delta )/4+1$. Let $L_j$ be an increasing compact exhaustion of $E$.

Using compact batching of a small open set, we recursively choose finite closed-arc unions $K_j\subset E$, of measure at most $3{\delta }_j$, such that $L_j$ is covered by the interiors of $K_1,\dots ,K_j$ andHere is the measure-theoretic justification. At step $j$, the uncovered portion has measure at most ${\delta }_j$. Its intersection with $L_j$ after removing the previously covered interiors is compact, and can be covered by finitely many closed arcs lying in $E$ with total length at most ${\delta }_j$ plus an arbitrarily small excess. By inner regularity of Lebesgue measure, a further compact subset of the uncovered open portion captures all but ${\delta }_{j+1}$ of its measure; cover it similarly. The combined finite arc union has length at most $3{\delta }_j$. This both covers the outstanding part of $L_j$ and ensures the next remainder estimate. In particular $E\subset {\bigcup }_j{K}_j$.

Apply the finite-arc construction to $K_j$ to get $P_j$ of degree $D_j$. Choose carriers $m_j$ recursively so that the bands $[m_j-D_j,m_j+D_j]$ are positive and strictly ordered with no overlap. SetThe frequency-separated Fourier block series and the Weierstrass M-test give a continuous complex-valued function with $\Vert f{\Vert }_{\infty }\le {\sum }_j{2}^{-j}=1$. At the cutoff $m_j$, all earlier bands have been included in full and every later band is absent. Thus on $K_j$,Every point of $E$ lies in some $K_j$, soValues of $\psi$ outside $0<\delta \le 1$ may be defined arbitrarily. The construction handles an arbitrary open set, not just one point or a finite collection of intervals. It also prevents cancellation explicitly: earlier bands contribute a bounded complete sum, and later bands contribute nothing at the chosen cutoff.

We prove the Kahane-Katznelson divergence theorem through an explicit small-norm block construction. Let $\mu$ be normalized Lebesgue measure on the circle.

First establish the compact-set Fourier amplification lemma. If a compact set $K$ satisfieswe can make a trigonometric polynomial $q$ with $\Vert q{\Vert }_{\infty }\le \epsilon$, supported in any sufficiently high interval of positive frequencies, whose partial prefix has magnitude greater than $M$ on $K$.

To construct it, choose a smooth nonnegative function $u$ equal to one near $K$, with values at most one and meanOuter regularity and a smooth cutoff give this choice; the stipulated bound on $\mu (K)$ leaves room between the two exponentials. The Schwarz integral on the unit diskhas positive real part in the disk, $H(0)=\delta$, and boundary real part $u$. Its holomorphic logarithmsatisfies $W(0)=0$ and $|\mathrm{Im}W|<\pi /2$. On $K$, the boundary value has $\mathrm{Re}W\ge \log (1/\delta )>4\pi M/\epsilon$. Smoothness of $u$ makes $H$ continuous at the boundary, and its positive boundary real part near $K$ makes $W$ continuous there.

Choose a radius just below one, then truncate the Taylor series of $W$ at that radius. This gives an analytic polynomial $R(e^{it})$ with zero constant term, degree $d$, andThe radial function is analytic beyond the closed unit disk, so the Taylor truncation is uniform on the whole circle. For $L>d$, putIts frequencies lie between $L-d$ and $L+d$, all positive. Its prefix through frequency $L$ includes exactly the negative-frequency half of $\mathrm{Im}R$ shifted into this interval:Consequently $|S_{L}q(t)|>\epsilon (4\pi M/\epsilon -1)/(2\pi )>M$ on $K$, whereas $\Vert q{\Vert }_{\infty }<\epsilon$. Increasing $L$ places the entire block above any previously used frequency.

We next use compact batching of a small open set to handle an arbitrary null set, without assuming that it is compact or a countable union of compact null sets. SetFor each $j$, choose an open $U_j\supseteq E$ with $\mu (U_j)<\exp (-8\pi j/{\epsilon }_{j,0})$. Decompose $U_j$ into countably many closed subarcs with pairwise disjoint interiors: subdivide each open component into closed pieces accumulating only at its excluded endpoints. Group these subarcs into finite successive batches $K_{j,k}$. After batch $k$, include enough pieces that the remaining total length is less than $\exp (-8\pi j/{\epsilon }_{j,k+1})$. Require each batch endpoint in the enumeration to increase. ThenEach batch is compact; endpoints shared by pieces have zero measure and do not affect the estimates.

Enumerate the pairs $(j,k)$ in diagonal order. Apply the block lemma with target $M=j$ to each $K_{j,k}$, and shift its spectrum above all preceding blocks. Denote the resulting polynomial by $q_{j,k}$ and setSince ${\sum }_{j,k}{\epsilon }_{j,k}=1/2$, this series is uniformly convergent and defines a continuous complex-valued function.

Fix $t\in E$. For every $j$ there is a $k$ with $t\in K_{j,k}$. The difference between the Fourier partial sum just before that block and the sum at its midpoint has magnitude greater than $j$: previous blocks cancel in the difference, and future blocks have not yet entered. As $j\to \infty$, these cutoffs tend to infinity. At least one of the two partial sums therefore has magnitude greater than $j/2$. The Fourier partial sums are unbounded, hence not Cauchy, at $t$. We have provedThis establishes divergence on every prescribed null set, including dense nonclosed null sets; it does not assert that the divergence set is exactly $E$.