Spring 2009 - Problem 7

Solution.

1. It suffices to prove the second claim, since it implies that for each compact $K$, $\abs{F\p{z}} < \infty$ almost everywhere on $K$, i.e., $F\p{z}$ converges almost everywhere on $K$, so $F\p{z}$ converges almost everywhere on $\C$.

   We may assume without loss of generality that $K = \cl{B\p{0, R}}$. Indeed, any compact set is contained within such a ball if $R$ is large enough, and we get the result by monotonicity.

   By applying Fubini-Tonelli, we may switch the order of integration in the following calculation:

   $$\begin{aligned} \int_{\cl{B\p{0,R}}} \abs{F\p{z}} \,\diff{A} &\leq \int_{\abs{z} \leq R} \int_{w \in \C} \frac{1}{\abs{z - w}} \,\diff\mu\p{w} \,\diff{A}\p{z} \\ &= \int_{w \in \C} \int_{\abs{z} \leq R} \frac{1}{\abs{z - w}} \,\diff{A}\p{z} \,\diff\mu\p{w} \\ &= \int_{\abs{w} \leq 2R} \int_{\abs{z} \leq R} \frac{1}{\abs{z - w}} \,\diff{A}\p{z} \,\diff\mu\p{w} + \int_{\abs{w} > 2R} \int_{\abs{z} \leq R} \frac{1}{\abs{z - w}} \,\diff{A}\p{z} \,\diff\mu\p{w} \\ &\leq \int_{\abs{w} \leq 2R} \int_{\abs{z} \leq R} \frac{1}{\abs{z - w}} \,\diff{A}\p{z} \,\diff\mu\p{w} + \int_{\abs{w} > 2R} \int_{\abs{z} \leq R} \frac{1}{R} \,\diff{A}\p{z} \,\diff\mu\p{w} \\ &\leq \int_{\abs{w} \leq 2R} \int_{\abs{z - w} \leq 3R} \frac{1}{\abs{z - w}} \,\diff{A}\p{z} \,\diff\mu\p{w} + \int_{\abs{w} > 2R} \int_{\abs{z} \leq R} \frac{1}{R} \,\diff{A}\p{z} \,\diff\mu\p{w} \\ &= \int_{\abs{w} \leq 2R} \int_0^{3R} \int_0^{2\pi} \frac{r}{r} \,\diff{r} \,\diff\theta \,\diff\mu\p{w} + \int_{\abs{w} > 2R} \pi R \,\diff\mu\p{w} \\ &= \int_{\abs{w} \leq 2R} 6\pi R \,\diff\mu\p{w} + \int_{\abs{w} > 2R} \pi R \,\diff\mu\p{w} \\ &= 6\pi R\mu\p{\abs{w} \leq 2R} + \pi R\mu\p{\abs{w} > 2R} \\ &< \infty. \end{aligned}$$

2. As before, it suffices to prove this for closed balls $\br{-R, R} \subseteq L$.

   Apply (1) and Fubini-Tonelli to $K = \br{-M, M} \times \br{n, n+1}$, where $n \in \Z$, $M \in \N$, which gives $\int_n^{n+1} \int_{-M}^M \abs{F\p{x + iy}} \,\diff{y} \,\diff{x} < \infty$. Thus, for almost every $y \in \br{n, n+1}$, we have $\int_{-M}^M \abs{F\p{x + iy}} \,\diff{x} < \infty$. Call this full measure set $Y_n^R$ and set

   $$Y_n = \bigcap_{M \in \N} Y_n^M.$$

   Notice that

   $$m\p{Y_n^\comp \cap \br{k, k+1}} \leq \sum_{k\in\Z} m\p{\p{Y_n^M}^\comp \cap \br{k, k+1}} = 0,$$

   so taking a union over $k \in \Z$, we see that $Y_n^\comp$ has Lebesgue measure zero. Moreover, if $y \in Y_n$, then $\int_{-M}^M \abs{F\p{x + iy}} \,\diff{x} < \infty$. Thus, if we set

   $$Y = \bigcup_{n\in\Z} Y_n,$$

   then $Y^\comp$ also has Lebesgue measure zero as a countable intersection of measure zero sets.

   Finally, let $R > 0$ and $y \in Y$. If $M \in \N$ is such that $R \leq M$, then by definition of $Y$,

   $$\int_{-R}^R \abs{F\p{x + iy}} \,\diff{x} \leq \int_{-M}^M \abs{F\p{x + iy}} \,\diff{x} < \infty,$$

   which proves the claim.

3. Notice that in (1), we proved that $\int_K \frac{1}{\abs{z - w}} \,\diff{A} < \infty$ for any compact set $K$. Thus, by replacing $\abs{F\p{z}}$ with $\int_\C \frac{1}{\abs{z - w}} \,\diff\mu\p{w}$ in (2), we see that for almost every horizontal line $L$ in $\C$ that

   $$\int_K \int_{w \in \C} \frac{1}{\abs{\p{x + iy} - w}} \,\diff\mu\p{w} \,\diff{x} < \infty$$

   for any compact $K \subseteq L$.

   If we run the same argument in (2) but with $x$ and $y$ switched, we obtain the analogous result for vertical lines with conull sets $X$ for vertical lines and $Y$ for horizontal ones.

   Let $S$ be the set of squares in $\C$ whose vertices in $X \times Y$, so $S$ contains almost every square in $\C$. Then in particular, we may integrate $\abs{F\p{z}}$ along $\partial S$. Combining this with our remark at the beginning, we may apply Fubini's theorem:

   $$\begin{aligned} \frac{1}{2\pi i} \int_{\partial S} F\p{z} \,\diff{z} &= \frac{1}{2\pi i} \int_{\partial S} \int_{w \in \C} \frac{1}{z - w} \,\diff\mu\p{w} \,\diff{z} \\ &= \int_{w \in \C} \p{\frac{1}{2\pi i} \int_{\partial S} \frac{1}{z - w} \,\diff{z}} \,\diff\mu\p{w} \\ &= \int_{w \in \C} \chi_S \,\diff\mu \\ &= \mu\p{S}, \end{aligned}$$

   where the second-to-last equality comes from Cauchy's integral formula, and this gives us the desired result.