GEORGIAN MATHEMATICAL JOURNAL: Vol. 3, No. 5, 1996, 447-456

ON THE TWO QUESTIONS OF LOHWATER AND PIRANIAN

M. GVARADZE

Abstract. The problem we are dealing with consists in the following: find the necessary and sufficient conditions for the zero measure subset of the circumference at which points the bounded analytic function has no radial limits.

1. For a function f : D → ℂ analytic and bounded in the unit disk D = {z : z ∈ ℂ, |z| < 1} and any point e^{iθ} ∈ C = {e^{iτ} : 0 ≤ τ < π} let f(e^{iθ}) = lim_{r→1} f(re^{iθ}) denote the radial limit of f. Fatou proved in 1906 that for such a function there exist radial limits except maybe for a set of points e^{iθ} of linear measure 0. Conversely, as Lusin showed in [1], for any set E of measure zero on C there exists a function analytic and bounded in D having no radial limits at the points of E.

Lohwater and Piranian noticed in [2] that “the set of nonexistence of radial limit is of second category for some bounded regular functions; it can even be a residual set on C; but we do not know of any case where a set E of second category, prescribed without reference to function theory, has been established as the precise set where the radial limit of some bounded regular function fails to exist.”

This is the first question we have to answer in this note. The second one is connected with the following statement (Theorem 8 in [2]).

Theorem. “Let the set E on C be of types F_{σ} and G_{δ} and of measure zero. Then there exists a function f(z), regular and bounded in D, which has the following properties: for each point e^{iθ} in E,

lim_{r→1} inf |f(re^{iθ})| = 0, \underline{lim}_{r→1} |f(re^{iθ})| = 1;

447 for each point e^{iθ} in C\E the radial limit f(e^{iθ}) exists and has modulus 1, except for a denumerable set of points where f(e^{iθ}) = 0.”

1991 Mathematics Subject Classification. 30B30.

Key words and phrases. Bounded analytic function, radial limit, modulus oscillation, Cantor set.

1072-947X/96/0900-0447$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1997 Plenum Publishing Corporation

In 1956 Lohwater and Piranian did not know “whether the converse of the theorem is true.”

Theorems 1 and 2 proved below give answers to these two questions.

2. To prove the theorems we need several lemmas. Some of them are of independent interest. In what follows

C(a, r) = {z : |z − a| ≤ r}, D(a, r) = {z : |z − a| < r};

the length of an interval I will be denoted by the same letter I.

Lemma 1. Let z_{0} = e^{it_{0}} and 0 < α < 1.

If z ∈ D(0, 1)\D(αz_{0}, 1 − α), then Re \frac{z_{0} + z}{z_{0} − z} ≤ \frac{α}{1 − α}.

If z ∈ C(αz_{0}, 1 − α), then Re \frac{z_{0} + z}{z_{0} − z} = \frac{α}{1 − α}.

If z ∈ D(αz_{0}, 1 − α), then Re \frac{z_{0} + z}{z_{0} − z} > \frac{α}{1 − α}.

If 0 < t < π/2, then

sup_{0≤r≤1} |\frac{1 + re^{it}}{1 − re^{it}} | < | cot \frac{t}{2}|.

For 0 < r < 1

|\frac{z_{0} + re^{it}}{z_{0} − re^{it}} − i cot \frac{t − t_{0}}{2} | =

= \frac{1 − r^{2}}{(1 − r)^{2} + 4r sin^{2} \frac{t−t_{0}}{2}} |1 − i\frac{1 − r}{1 + r} cot \frac{t − t_{0}}{2} |.

Proof. The linear-fractional mapping \frac{z_{0}+z}{z_{0}−z} maps the circle C(αz_{0}, 1 − α) on the straight line x = α/(1 − α) and the disk D(αz_{0}, 1 − α) on the halfplane {z : Re z > α/(1 − α)} which imply the first three assertions of the lemma. To prove the next inequality note that the straight line containing the radius \{re^{it} : 0 ≤ r ≤ 1\} is mapped by the mapping \frac{1+z}{1−z} on the circle having the center on the imaginary axis and intersecting this axis at the points −i tan \frac{t}{2} and a = i cot \frac{t}{2}. The radius itself is mapped onto the smaller arc of this circle (lying in the first quadrant) with the endpoints 1 and b = \frac{1+e^{it}}{1−e^{it}}. Simple analysis of the triangle with vertices 0, a, and b shows that |a| > |b|.

The last assertion is proved as follows (τ = t − t_{0}):

\frac{z_{0} + re^{it}}{z_{0} − re^{it}} − i cot \frac{τ}{2} = \frac{1 + re^{iτ}}{1 − re^{iτ}} − i cot \frac{τ}{2} =

= \frac{1 − r^{2}}{(1 − r)^{2} + 4r sin^{2} \frac{τ}{2}} + i\frac{2r sin τ}{(1 − r)^{2} + 4r sin^{2} \frac{τ}{2}} − i cot \frac{τ}{2} =

= \frac{1 − r^{2}}{(1 − r)^{2} + 4r sin^{2} \frac{τ}{2}} (1 − i\frac{1 − r}{1 + r} cot \frac{τ}{2} ).

Lemma 2. ρ(e^{it}, C(α, 1 − α)) ≥ 2α sin^{2} \frac{t}{2}, where ρ(E, F) is a distance between the sets E and F, and α ∈ (0, \frac{1}{2}).

Proof. It is evident that ρ(e^{it}, C(α, 1 − α)) = |e^{it} − α| − (1 − α) =

= \sqrt{(1 − α)^{2} + 4α sin^{2} \frac{t}{2}} − (1 − α) ≥ 2α sin^{2} \frac{t}{2} (equality holds if t = kπ, k ∈ ℤ).

Lemma 3. Suppose θ_{n} = arg z_{n} and ∑_{n=1}^{∞} I_{n}| cot \frac{θ−θ_{n}}{2} | < ∞. Then the

function g(z) = ∑_{n=1}^{∞} I_{n} \frac{z_{n}+z}{z_{n}−z} has a radial limit of modulus one at the point e^{iθ}.

Proof. By Lemma 1, | ∑_{n=1}^{∞} I_{n} \frac{z_{n}+z}{z_{n}−z} | ≤ ∑_{n=1}^{∞} I_{n}| cot \frac{θ−θ_{n}}{2} |. Therefore g is continuous on the radius [0, e^{iθ}].

Let d(E, F) denote an arcdistance between the subsets E and F of C and ∂G be a boundary of G.

Lemma 4. Let G ⊂ G_{1} ⊂ G_{0} be open subsets of the circle. Suppose

G = ∪^{∞}_{n=1}I_{n}, G_{1} = ∪^{∞}_{n=1}J_{n} and z_{n} ∈ \overline{I_{n}}. If e^{iθ} ∉ G_{0} and d(J_{k}, ∂G_{0}) > 2^{k}J_{k},

then there exists lim_{r→1} ∑_{n=1}^{∞} I_{n} \frac{z_{n}+re^{iθ}}{z_{n}−ze^{iθ}}.

Proof. If I_{n} ⊂ J_{k}, then

sup_{0≤r≤1} |\frac{z_{n} + re^{iθ}}{z_{n} − re^{iθ}} | ≤ | cot \frac{θ − θ_{n}}{2} | ≤ cot \frac{d(e^{iθ}, J_{k})}{2},

and therefore

∑_{k=1}^{∞} ∑_{I_{n}⊂J_{k}} |I_{n} \frac{z_{n} + re^{iθ}}{z_{n} − re^{iθ}} | ≤ ∑_{k=1}^{∞} cot \frac{d(e^{iθ}, J_{k})}{2} ∑_{I_{n}⊂J_{k}} I_{n} ≤

≤ ∑_{k=1}^{∞} J_{k}| cot \frac{d(e^{iθ}, J_{k})}{2}| < ∞.

Using Lemma 3, we can prove Lemma 4.

The following lemma is the basic one.

450 M. GVARADZE

Lemma 5. Suppose G= ∪^{∞}_{n=1}I_{n} to be an open set. Let z_{n} ∈I_{n}, d(z_{n}, ∂I_{n})> I_{n}/4 and let

∑_{n∉\{n_{k}\}} \frac{I_{n}}{d(e^{iθ}, I_{n})} < ∞, lim_{k→∞} \frac{I_{n_{k+1}}}{I_{n_{k}}} = 0.

Then the function g(z) = ∑_{n=1}^{∞} I_{n} \frac{z_{n}+z}{z_{n}−z} has radial limit at the point e^{iθ} if and only if the series

∑_{n=1}^{∞} I_{n} cot \frac{θ − arg z_{n}}{2}  (1)

is convergent. Proof. By Lemma 3 since the function ∑_{n∉\{n_{k}\}} I_{n} \frac{z_{n}+z}{z_{n}−z} has radial limit at the point e^{iθ}, then without loss of generality we may assume that

lim_{n→∞} \frac{I_{n+1}}{I_{n}} = 0, \frac{I_{n+1}}{I_{n}} ≤ \frac{1}{2}.

Denote ρ_{n} = 1 − I_{n+1} and suppose that ρ_{N−1} < |z| ≤ ρ_{N}. We have

g(z) = ∑_{n=1}^{N−1} I_{n}(\frac{z_{n} + z}{z_{n} − z} − i cot \frac{θ − arg z_{n}}{2} ) + I_{N} \frac{z_{N} + z}{z_{N} − z} +
+ I_{N+1} \frac{z_{N+1} + z}{z_{N+1} − z} + ∑_{N+2}^{∞} I_{n} \frac{z_{n} + z}{z_{n} − z} + i ∑_{n=1}^{N−1} I_{n} cot \frac{θ − arg z_{n}}{2} .  (2)

If n ≥ N + 2, then

I_{n}|\frac{z_{n} + z}{z_{n} − z} | ≤ 2\frac{I_{n}}{1 − |z|} ≤ 2\frac{I_{n}}{I_{N+1}} ≤ 2^{N−n} \frac{I_{N+2}}{I_{N+1}} ,

whence

| ∑_{n=N+2}^{∞} I_{n} \frac{z_{n} + z}{z_{n} − z} | ≤ \frac{I_{N+2}}{I_{N+1}} .  (3)

By Lemma 1

I_{n}|\frac{z_{n} + z}{z_{n} − z} − i cot \frac{θ − arg z_{n}}{2} | ≤ 4π^{3}I_{n} \frac{1 − r}{(θ − arg z_{n})^{2}} ≤ 16π^{3} \frac{I_{N}}{I_{n}} .

Therefore

| ∑_{n=1}^{N−1} I_{n}(\frac{z_{n} + z}{z_{n} − z} − i cot \frac{θ − arg z_{n}}{2} )| ≤ 16π^{3} \frac{I_{N}}{I_{N−1}} .  (4)

If lim_{n→∞} I_{n} cot \frac{θ−arg z_{n}}{2} = 0, then by Lemma 1, lim_{n→∞} I_{n} \frac{z_{n}+z}{z_{n}−z} = 0. Thus, by (3) and (4) the right-hand side of (2) has radial limit if and only if (1) is convergent.

Finally, if \overline{lim}_{n→∞} I_{n}| cot \frac{θ−arg z_{n}}{2} | ≥ δ > 0, we have

Re I_{N+1} \frac{z_{N+1} + ρ_{N}e^{iθ}}{z_{N+1} − ρ_{N}e^{iθ}} ≥ I_{N+1} \frac{1 − ρ_{N}^{2}}{(1 − ρ_{N})^{2} + 4ρ_{N} sin^{2} \frac{θ−θ_{N}}{2}} ≥

≥ I_{N+1} \frac{1 − ρ_{N}}{(1 − ρ_{N})^{2} + 4\frac{I_{N+1}^{2}}{δ^{2}}} = \frac{I_{N+1}^{2}}{I_{N+1}^{2} + \frac{4}{δ^{2}} I_{M+1}^{2}} ,

since sin^{2} \frac{θ−θ_{N+1}}{2} ≤ \frac{I_{N+1}^{2}}{δ^{2}} , where θ_{N} = arg z_{N+1};

Re I_{N} \frac{z_{N} + ρ_{N}e^{iθ}}{z_{N} − ρ_{N}e^{iθ}} ≤ I_{N} \frac{2(1 − ρ_{N})}{(1 − ρ_{N})^{2} + 4ρ_{N} sin^{2} \frac{θ−θ_{N}}{2}} ≤

≤ \frac{2I_{N}I_{N+1}}{4ρ_{N}(\frac{2}{π} \frac{θ−θ_{N}}{2} )^{2}} ≤ \frac{2I_{N}I_{N+1}}{4\frac{1}{2} \frac{1}{π^{2}} (\frac{1}{4}I_{N})^{2}} = 16π^{2} \frac{I_{N+1}}{I_{N}} ,

since |θ − θ_{N}| ≥ \frac{1}{4}I_{N}, e^{iθ_{N}} ∉ I_{N}, ρ_{N} ≥ \frac{1}{2}.

By the notation ρ′_{N} = 1 − \sqrt{I_{N}I_{N+1}} we get

Re I_{N+1} \frac{z_{N+1} + ρ′_{N}e^{iθ}}{z_{N+1} − ρ′_{N}e^{iθ}} ≤ I_{N+1} \frac{2(1 − ρ′_{N})}{(1 − ρ′_{N})^{2}} ≤

≤ 2\frac{I_{N+1}}{1 − ρ′_{N}} = 2\frac{I_{N+1}}{\sqrt{I_{N}I_{N+1}}} = 2\sqrt{\frac{I_{N+1}}{I_{N}}}

and

Re I_{N} \frac{z_{N} + ρ′_{N}e^{iθ}}{z_{N} − ρ′_{N}e^{iθ}} ≤ I_{N} \frac{2(1 − ρ′_{N})}{(I_{N}^{2})/(16π^{2})} =

= 32π^{2} \frac{I_{N} \sqrt{I_{N}I_{N+1}}}{I_{N}^{2}} = 32π^{2}\sqrt{\frac{I_{N+1}}{I_{N}}} .

Taking into acount these inequalities we can conclude that the right-hand side of (2) has no radial limit.

Definition 1. We say that E is an arrangeable set if there exist a countable set (z_{mn}) = (e^{iθ_{mn}}), a sequence of intervals (arcs) (I_{mn}), and a sequence of positive numbers (α_{k}) satisfying the following conditions:

1. z_{mn} ∈ I_{mn}, 4d(z_{mn}, ∂I_{mn}) > I_{mn};

2. lim_{n→∞} \frac{I_{m(n+1)}}{I_{mn}} = 0, m = 1, 2, …;

3. I_{(m+1)n} ⊂ ∪^{∞}_{i=1}(I_{mi}\{z_{mi}}), m = 1, 2, …;

4. \sqrt{I_{mn}} < 2^{−n}α_{m}α_{m+1}d(I_{mn}, ∂G_{m−1}), m > 1, where G_{m} = ∪^{∞}_{n=1}(I_{mn}\

{z_{mn}}), a_{k} ↘ 0, ∑ α_{k} < ∞;

5. E = ( ∩^{∞}_{m=1}G_{m}) ∪ ( ∪^{∞}_{m=1}Q_{m}), where

Q_{m} = \{e^{iθ} : e^{iθ} ∉ G_{m}, ∑_{n=1}^{∞} I_{mn} cot \frac{θ − θ_{mn}}{2} is divergent\}.

Lemma 6. Let E be an arrangeable set. Then the function f(z) =

exp\{−g(z)\}, with g(z) = ∑_{m=1}^{∞} g_{m}(z), g_{m}(z) = \frac{1}{α_{m}} ∑_{n=1}^{∞} I_{mn} \frac{z_{mn}+z}{z_{mn}−z}, has no

radial limit on the set E only.

Proof. Since

∑_{m=1}^{∞} \frac{1}{α_{m}} ∑_{n=1}^{∞} I_{mn} ≤ ∑_{m=1}^{∞} \frac{1}{α_{m}} ∑_{n=1}^{∞} 2^{−2n}α_{m}^{2}α_{m+1}^{2} < ∞,

we conclude that f is analytic in the unit disk.

Suppose e^{iθ} ∉ G_{1}. We have

∑_{m=2}^{∞} \frac{1}{α_{m}} ∑_{n=1}^{∞} I_{mn}| cot \frac{θ−θ_{mn}}{2} |< ∑_{m=2}^{∞} \frac{1}{α_{m}} ∑_{n=1}^{∞} I_{mn} \frac{1}{d(I_{mn}, ∂G_{m−1})} <

< ∑_{m=2}^{∞} \frac{1}{α_{m}} ∑_{n=1}^{∞} 2^{−4n}α_{m}^{2}α_{m+1}^{2}d(I_{mn}, ∂G_{m−1}) < ∑_{m=2}^{∞} α_{m}α_{m+1}^{2} < ∞.

Hence by Lemma 3, g − g_{1} has radial limit at e^{iθ}, and thus by our basic lemma we can conclude that g has no radial limit only at Q_{1}.

Now if e^{iθ} ∈ G_{m}\G_{m+1}, m = 1, 2, …, then

∑_{k=1}^{m} \frac{1}{α_{k}} ∑_{n=1}^{∞} I_{kn}| cot \frac{θ − θ_{kn}}{2} | < \frac{1}{d(e^{iθ}, ∂G_{m})} ∑_{k=1}^{m} \frac{1}{α_{k}} ∑_{n=1}^{∞} I_{kn} < ∞,

and

∑_{k=m+2}^{∞} \frac{1}{α_{k}} ∑_{n=1}^{∞} I_{kn}| cot \frac{θ − θ_{kn}}{2} | <

< ∑_{k=m+2}^{∞} \frac{1}{α_{k}} ∑_{n=1}^{∞} I_{kn} \frac{1}{d(I_{kn}, ∂G_{k−1})} < ∑_{k=m+2}^{∞} α_{k}α_{k+1}^{2} < ∞.

Therefore arguing as above we come to the conclusion that g has no radial limit at Q_{m+1}. Suppose now that e^{iθ} ∈ ∩^{∞}_{m=1}G_{m}, and denote O_{m} =

∪^{∞}_{n=1}D(α_{m}^{2}z_{mn}, 1 − α_{m}^{2}), C_{m} = ∂O_{m}.

Let z ∈ C_{m} and k ≤ m; then since z ∉ D(α_{m}^{2}z_{mn}, 1 − α_{m}^{2}), we have

Re g_{k}(z) ≤ \frac{1}{α_{k}} ∑_{n=1}^{∞} I_{kn} \frac{α_{m}^{2}}{1 − α_{m}^{2}} ≤ \frac{2α_{m}^{2}}{α_{k}} μG_{k} ≤ α_{m}.

Further, if z ∈ C_{m} and k > m, then according to Lemma 2,

Re \frac{z_{kn} + z}{z_{kn} − z} ≤ \frac{2}{ρ(z_{kn}, C_{m})} < \frac{1}{α_{m}^{2} sin^{2} \frac{θ_{kn}−θ_{mi}}{2}} < \frac{π^{2}}{α_{m}^{2}d^{2}(I_{kn}, ∂G_{k−1})} .

Hence

Re g_{k}(z) ≤ \frac{1}{α_{k}} ∑_{n=1}^{∞} 2^{−2n}α_{k}^{2}α_{k+1}^{2}d^{2}(I_{kn}, ∂G_{k−1})\frac{π^{2}}{α_{m}^{2}d^{2}(I_{kn}, ∂G_{k−1})} ≤

≤ \frac{α_{k}α_{k+1}^{2}}{α_{m}^{2}} ≤ α_{k+1}.

Thus, if z ∈ C_{m}, then

Re g(z) ≤ ∑_{k=1}^{m} α_{m} + ∑_{k=m+1}^{∞} α_{k} = mα_{m} + ∑_{k=m+1}^{∞} α_{k},

whence

\underline{lim}_{r→1} Re g(re^{iθ}) = 0.  (5)

Since e^{iθ} ∈ G_{m}, there exists an integer j such that e^{iθ} ∈ I_{mj}. It is clear that the radius [0, e^{iθ}] intersects the circle C((1 − I_{mj})z_{mj}, I_{mj}) in which

Re g(re^{iθ}) ≥ Re g_{m}(re^{iθ}) ≥ Re \frac{1}{α_{m}} I_{mj} \frac{z_{mj} + z}{z_{mj} − z} ≥ \frac{1}{2α_{m}} .

Hence, \overline{lim}_{r→1} Re g(re^{iθ}) = ∞, which together with (5) gives

\underline{lim}_{r→1} |f(re^{iθ})| = 0, \overline{lim}_{r→1} |f(re^{iθ})| = 1.

3. Let us now formulate and prove our theorems.

Theorem 1. There exists an arrangeable set of second category.

Proof. Let Q = {x_{n}} be a countable subset of a unit circle and G_{0} be an open subset covering Q and μG_{0} < 2π (μ denotes the Lebesgue measure).

Cover x_{1} by the interval J_{1} such that ∂J_{1} ∩ Q = 0 and \sqrt{J_{1}} < 2^{−1}α_{1}α_{2} d(J_{1}, ∂G_{0}). Let x_{k_{2}} be the first element of the sequence Q not belonging to J_{1}. Cover x_{k_{2}} by the interval J_{2} such that ∂J_{2} ∩ Q = 0, J_{1} ∩ J_{2} = ∅, \sqrt{J_{2}} < 2^{−2}α_{1}α_{2}d(J_{2}, ∂G_{0}), \frac{J_{2}}{J_{1}} < \frac{1}{2}.

Given J_{1}, J_{2}, …, J_{n−1}, let x_{k_{n}} be the first element of Q not belonging

to ∪^{n−1}_{k=1}J_{k}. Cover x_{k_{n}} by the interval J_{n} such that ∂J_{n} ∩ Q = ∅, J_{n} ∩

( ∪^{n−1}_{k=1}G_{k}) = ∅, \sqrt{J_{n}} < 2^{−n}α_{1}α_{2}d(J_{n}, ∂G_{0}), \frac{J_{n}}{J_{n−1}} < \frac{1}{n}. Select z_{1n} such that

z_{1n} ∈ J_{n}, 4d(z_{1n}, ∂J_{n}) > J_{n} and z_{1n} ∉ Q. Denote I_{1n} = J_{n}\{z_{1n}} and

G_{1} = ∪^{∞}_{n=1}I_{1n}.

Taking G_{1} instead of G_{0} and arguing analogously, we will obtain new sequences of the intervals J_{n} and points z_{2n} with the following properties:

(a) z_{2n} ∈ J_{n}, z_{2n} ∉ Q;

(b) J_{n} ∩ J_{k} = ∅, n ≠ k;

(c) 4d(z_{2n}, J_{n}) > J_{n};

(d) \sqrt{J_{n}} < 2^{−n}α_{2}α_{3}d(J_{n}, ∂G_{1});

(e) lim_{n→∞} \frac{J_{n+1}}{J_{n}} = 0.

Denote I_{2n} = J_{n}\{z_{2n}} and G_{2} = ∪^{∞}_{n=1}I_{2n}.

Repeating the above process for G_{2}, G_{3}, etc., we shall get the sequences z_{mn} and I_{mn} which define the arrangeable set E. If Q is a countable everywhere dense subset, then the obtained set E will be of second category.

Theorem 2. There exist a set E and a bounded analytic function f with the following conditions:

(a) E is of G_{δ} type;

(b) E is not of F_{σ} type;

(c) if e^{iθ} ∈ E, then \underline{lim}_{r→1} |f(re^{iθ})| = 0, \overline{lim}_{r→1} |f(re^{iθ})| = 1;

(d) the function f on C\E has radial limit of modulus one except a countable set where the radial limit is zero.

Proof. Consider the Cantor set E without ends on its adjacent intervals. It is clear that E is of type G_{δ} and not of type F_{σ}.

Suppose E = ∩^{∞}_{k=1}G_{k} and ∑_{k=1}^{∞} α_{k} < 1. Without loss of generality we may

assume that the ends of component intervals I_{1n} of G_{1} belong to adjacent intervals of Cantor’s set. Cover every I_{1n} ∩ \overline{E} by an open set H_{1n} such that \sqrt{μH_{1n}} < 2^{−n}α_{1}α_{2}d(H_{1n}, ∂I_{1n}), where H_{1n} is a finite union of intervals. Put R_{1} = ∪^{∞}_{n=1}H_{1n} = ∪^{∞}_{k=1}J_{1k}. Suppose z_{1k} is the right end of the interval J_{1k}.

If e^{iθ} ∉ G_{1}, we have

∑_{k=1}^{∞} J_{1k}| cot \frac{θ − θ_{1k}}{2} | ≤ ∑_{n=1}^{∞} ∑_{J_{1k}⊂I_{1n}} J_{1k} \frac{π}{d(H_{1n}, ∂I_{1n})} ≤

≤ π ∑_{n=1}^{∞} \frac{μH_{1n}}{d(H_{1n}, ∂I_{1n})} ≤ πα_{1}α_{2}

and if e^{iθ} ∈ I_{1n}, then

∑_{J_{1k}∩I_{1n}=∅} J_{1k}| cot \frac{θ − θ_{1k}}{2} | ≤ \frac{π}{d(e^{iθ}, ∂I_{1n})} ∑_{n=1}^{∞} μH_{1n} = \frac{πμR_{1}}{d(e^{iθ}, ∂I_{1n})}.

Hence, according to Lemma 3 and the fact that intervals contained in I_{1n} are finite, we conclude that f_{1}(z) = exp{−g_{1}(z)}, and

g_{1}(z) = \frac{1}{α_{1}} ∑_{k=1}^{∞} J_{1k} \frac{z_{1k} + z}{z_{1k} − z}

has everywhere a radial limit. Put R_{1} ∩G_{2} = ∪^{∞}_{n=1}I_{2n}. We assume again that the ends of I_{2n} belong to the adjacent intervals of Cantor’s set. Cover every I_{2n} ∩ \overline{E} by open sets H_{2n} such that \sqrt{μH_{2n}} < 2^{−n}α_{2}α_{3}d(H_{2n}, ∂I_{2n}), where H_{2n} is a finite union of intervals. Put R_{2} = ∪^{∞}_{n=1}H_{2n} = ∪^{∞}_{k=1}J_{2k}. Suppose z_{2k} is the right end of J_{2k}.

If e^{iθ} ∉ R_{1} ∩ G_{2}, we have

∑_{k=1}^{∞} J_{2k}| cot \frac{θ − θ_{2k}}{2} | ≤ πα_{2}α_{3}

and if e^{iθ} ∈ I_{2n}, then

∑_{J_{2k}∩I_{2n}=∅} J_{2k}| cot \frac{θ − θ_{2k}}{2} | ≤ \frac{πμR_{2}}{d(e^{iθ}, ∂I_{2n})} .

Hence, according to Lemma 3 and the fact that intervals contained in I_{2n} are finite, we conclude that f_{2}(z) = exp{−g_{2}(z)}, and

g_{2}(z) = \frac{1}{α_{2}} ∑_{k=1}^{∞} J_{2k} \frac{z_{2k} + z}{z_{2k} − z}

has everywhere a radial limit.

Given R_{1}, R_{2}, …, R_{m−1}, put R_{m−1} ∩ G_{m} = ∪^{∞}_{n=1}I_{mn}. Assume that the

ends of I_{mn} belong to adjacent intervals of Cantor’s set. Cover every I_{mn}∩\overline{E} by open sets H_{mn} such that \sqrt{μH_{mn}} < 2^{−n}α_{m}α_{m+1}d(H_{mn}, ∂I_{mn}), where H_{mn} is a finite union of intervals. Put R_{m} = ∪^{∞}_{n=1}H_{mn} = ∪^{∞}_{k=1}J_{mk}. Suppose z_{mk} is the right end of J_{mk}.

If e^{iθ} ∉ R_{m−1} ∩ G_{m}, we have

∑_{k=1}^{∞} J_{mk}| cot \frac{θ − θ_{mk}}{2} | ≤ πα_{m}α_{m+1}  (6)

and if e^{iθ} ∈ I_{mn}, then

∑_{J_{mk}∩I_{mn}=∅} J_{mk}| cot \frac{θ − θ_{mk}}{2} | ≤ \frac{πμR_{m}}{d(e^{iθ}, ∂I_{mn})} .

The function f_{m}(z) = exp{−g_{m}(z)} with

g_{m}(z) = \frac{1}{α_{m}} ∑_{k=1}^{∞} J_{mk} \frac{z_{mk} + z}{z_{mk} − z}

has everywhere a radial limit.

Put

f(z) = ∏_{m=1}^{∞} f_{m}(z) = exp { − ∑_{m=1}^{∞} g_{m}(z)}.

It is evident that if e^{iθ} = z_{mk}, then lim_{r→1} f(re^{iθ}) = 0. Since R_{m} ⊂ G_{m} and

R_{m} ⊃ E, then ∩^{∞}_{m=1}R_{m} = E.

If e^{iθ} ∉ E, then there exists an integer n such that e^{iθ} ∉ R_{n}. By (6) we have

∑_{m=n}^{∞} \frac{1}{α_{m}} ∑_{k=1}^{∞} J_{mk}| cot \frac{θ − θ_{mk}}{2} | ≤ π ∑_{m=n}^{∞} α_{m+1} < ∞.

Therefore, by Lemma 3, Re ∑_{m=n}^{∞} g_{m}(re^{iθ}) has zero radial limit.

Suppose finally e^{iθ} ∈ E. Then using the definitions and arguing as in Lemma 6, we may conclude that

\underline{lim}_{r→1} |f(re^{iθ})| = 0, \overline{lim}_{r→1} |f(re^{iθ})| = 1.

References

1. N. N. Lusin, Sur la representation conforme. Ivanovo, Izv. Politekh. Instituta 2(1919), 77–80.

2. A. J. Lohwater and G. Piranian, The boundary behavior of functions analytic in a disk. Ann. Acad. Sci. Fenn., Ser. A.I 239(1957), 1–17.

(Received 28.10.1993; revised version 30.11.1994)

Author’s address: A. Razmadze Mathematical Institute of Georgian Academy of Sciences 1, M. Alexidze St., Tbilisi 380093, Republic of Georgia