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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.