Let $(X,\mathscr M,\mu)$ be a measure space and $(Y,\|\cdot\|)$ a separable Banach space with $\{y_n\}_{n=1}^{\infty}$ being a dense subset in it. Suppose that $f:X\to Y$ is a Borel measurable function and $$\int \|f(x)\|\,\mathrm d\mu<+\infty.$$ Suppose also that the sets $(E_{nj})_{n,j=1}^{\infty}$ in $X$ are such that

  • all of them are in $\mathscr M$ and their measures are finite;
  • for any given $j\in\mathbb Z_+$, $(E_{nj})_{n=1}^{\infty}$ are disjoint and their union gives $\{x\in X\,|\,f(x)\neq 0\}$;
  • if $x\in E_{nj}$, then $\|y_n-f(x)\|<1/j\cdot\|y_n\|$.

What I want to show is that the set $$\bigcap_{\ell=1}^{\infty}\bigcup_{k=\ell}^{\infty}\bigsqcup_{n=k+1}^{\infty}E_{nk}$$ is of measure zero (the notation $\sqcup$ emphasizes that the union is one of disjoint sets). I'm afraid I cannot use a limiting argument that exploits measures being continuous from above, as the sets $$\bigcup_{k=\ell}^{\infty}\bigsqcup_{n=k+1}^{\infty}E_{nk}$$ may be of infinite measure for all $\ell\in\mathbb Z_+$. In particular, if $\|f(x)\|>0$ for all $x\in X$ and $\mu(X)=+\infty$, then this fear is justified, as $X=\bigcup_{n=1}^{\infty} E_{nk}$ for all $k\in\mathbb Z_+$ by the second assumption above.

If it is of any help, we know also that for any $\varepsilon>0$, the following holds: $$Y\setminus\{0\}\subseteq\bigcup_{n=1}^{\infty}\left\{y\in Y\,\big|\,\|y_n-y\|<\varepsilon\|y_n\|\right\}.$$

More generally (it would be enough for my purposes), is it possible to choose a map from $\mathbb Z_+$ to itself $k\mapsto N_k$ such that $$\bigcap_{\ell=1}^{\infty}\bigcup_{k=\ell}^{\infty}\bigsqcup_{n=N_k+1}^{\infty}E_{nk}$$ is of measure zero?

Any suggestion will be dearly appreciated.

  • $\begingroup$ With $1/j \times \Vert y_n \Vert$ you mean $\Vert y_n \Vert /j$? $\endgroup$ Commented Jan 3, 2014 at 18:10
  • $\begingroup$ @EmanuelePaolini Yes, it's ordinary division. I changed the $\times$ to $\cdot$ for clarity. $\endgroup$
    – triple_sec
    Commented Jan 3, 2014 at 18:11
  • $\begingroup$ Is $\mathscr{M}$ $\sigma$-finite? $\endgroup$
    – Norbert
    Commented Jan 3, 2014 at 20:41
  • $\begingroup$ @Norbert Not necessarily. The only finiteness assumptions made concern the measures of the sets $(E_{nj})_{n,j=1}^{\infty}$ and the integral of $\|f(x)\|$. $\endgroup$
    – triple_sec
    Commented Jan 3, 2014 at 20:43
  • $\begingroup$ @Norbert Addition: By the second assumption, the set on which $f$ does not vanish is actually $\sigma$-finite. $\endgroup$
    – triple_sec
    Commented Jan 5, 2014 at 12:07

1 Answer 1


I got it. Let $K_0\equiv\{x\in X\,|\,f(x)\neq 0\}$. By the second assumption, we have that $$\bigsqcup_{n=1}^{\infty}E_{nk}=K_0\quad\forall k\in\mathbb Z_+.$$ Moreover, $\mu(E_{n1})<+\infty$ for all $n\in\mathbb Z_+$, so let $$t_n\equiv\max_{m\in\{1,\ldots,n\}}\left\{\mu(E_{m1})\right\}+n\quad\forall n\in\mathbb Z_+.$$ It is easy to see that $t_{n+1}>t_n\geq n$ and $\mu(E_{n1})\leq t_n$ for all $n\in\mathbb Z_+$. Define $\mu_1:\mathscr M\to[0,+\infty]$ as follows: $$\mu_1(E)\equiv\sum_{n=1}^{\infty}\frac{1}{2^{t_n}}\mu(E\cap E_{n1})\quad\forall E\in\mathscr M.$$

$\textbf{Claim 1:}\quad$ $\mu_1$ is a finite measure.

$\textit{Proof:}\quad$ Clearly, $$\mu_1(\varnothing)=\sum_{n=1}^{\infty}\frac{1}{2^{t_n}}\mu(\varnothing\cap E_{n1})=\sum_{n=1}^{\infty}\frac{1}{2^{t_n}}\times0=0.$$ Now, let $$E\equiv\bigsqcup_{j=1}^{\infty} E_j,\quad\text{where}\{E_j\}_{j=1}^{\infty}\subseteq\mathscr M.$$ Then, \begin{align*} \mu_1(E)=&\,\sum_{n=1}^{\infty}\frac{1}{2^{t_n}}\mu\left(\bigsqcup_{j=1}^{\infty}(E_j\cap E_{n1})\right)=\sum_{n=1}^{\infty}\frac{1}{2^{t_n}}\sum_{j=1}^{\infty}\mu(E_j\cap E_{n1})=\sum_{n=1}^{\infty}\sum_{j=1}^{\infty}\frac{1}{2^{t_n}}\mu(E_j\cap E_{n1})\\=&\,\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{2^{t_n}}\mu(E_j\cap E_{n1})=\sum_{j=1}^{\infty}\mu_1(E_j). \end{align*} The third and fourth equalities need further justification. First, we have, for all $n\in\mathbb Z_+$, that \begin{align*} \sum_{j=1}^{\infty}\mu(E_j\cap E_{n1})=\mu(E\cap E_{n1})\leq\mu(E_{n1})<+\infty, \end{align*} so that moving the term $1/2^{t_n}$ inside the sum is legitimate, indeed. Second, \begin{align*} &\,\sum_{n=1}^{\infty}\sum_{j=1}^{\infty}\frac{1}{2^{t_n}}\mu(E_j\cap E_{n1})=\sum_{n=1}^{\infty}\frac{1}{2^{t_n}}\mu(E\cap E_{n1})\leq\sum_{n=1}^{\infty}\frac{1}{2^{t_n}}\mu(E_{n1})\\ \leq&\,\sum_{n=1}^{\infty}\frac{t_n}{2^{t_n}}\leq\sum_{n=1}^{\infty}\frac n{2^n}=2. \end{align*} The penultimate inequality follows from the facts that $t_n\geq n$ for all $n\in\mathbb Z_+$ and that the mapping $n\mapsto n/2^n$ is non-increasing, which is not difficult to show (nor is the fact that the last series converges to $2$). Hence, this non-negative double sum is convergent, so that the order of summation can be changed, indeed.

Finally, this argument also reveals that $\mu_1(X)\leq 2$. $\quad\blacksquare$

$\textbf{Claim 2:}\quad$ If $E\in\mathscr M$, $E\subseteq K_0$, and $\mu_1(E)=0$, then $\mu(E)=0$.

$\textit{Proof:}\quad$ If $E$ satisfies the hypotheses, then we have that $$E=E\cap K_0=\bigsqcup_{n=1}^{\infty}E\cap E_{n1}.$$ That $\mu_1(E)=0$ clearly implies that $\mu(E\cap E_{n1})=0$ for all $n\in\mathbb Z_+$. Therefore, $$\mu(E)=\sum_{n=1}^{\infty}\mu(E\cap E_{n1})=0.\quad\blacksquare$$

Now, let $$S\equiv\bigcap_{\ell=1}^{\infty}\bigcup_{k=\ell}^{\infty}\bigsqcup_{n=N_k+1}^{\infty}E_{nk},$$ where $k\mapsto N_k$ is a mapping to be characterized later. Clearly, $S\subseteq K_0$, for suppose that $x\in S$ and $f(x)=0$. Then, $x\in E_{nk}$ for some $n,k\in\mathbb Z_+$, so that, by the third hypothesis, $$\|y_n\|=\|y_n-f(x)\|<\frac{1}{k}\|y_n\|\leq\|y_n\|,$$ a contradiction.

Fix $k\in\mathbb Z_+$. Since $$\mu_1(K_0)=\mu_1\left(\bigsqcup_{n=1}^{\infty}E_{nk}\right)=\sum_{n=1}^{\infty}\mu_1(E_{nk})<+\infty,$$ there exists some $N_k\in\mathbb Z_+$ such that $$\sum_{n=N_k+1}^{\infty}\mu_1(E_{nk})=\mu_1\left(\bigsqcup_{n=N_k+1}^{\infty} E_{nk}\right)<\frac{1}{2^k}.$$ Then, we have that for all $\ell\in\mathbb Z_+$, $$\mu_1\left(\bigcup_{k=\ell}^{\infty}\bigsqcup_{n=N_k+1}^{\infty} E_{nk}\right)\leq\sum_{k=\ell}^{\infty}\mu_1\left(\bigsqcup_{n=N_k+1}^{\infty} E_{nk}\right)\leq\sum_{k=\ell}^{\infty}\frac1{2^k}=\frac{1}{2^{\ell-1}}<+\infty.$$ Since $$\bigcup_{k=\ell}^{\infty}\bigsqcup_{n=N_k+1}^{\infty} E_{nk}\supseteq\bigcup_{k=\ell+1}^{\infty}\bigsqcup_{n=N_k+1}^{\infty} E_{nk}\quad\forall\ell\in\mathbb Z_+,$$ we have that $$\mu_1(S)=\lim_{\ell\to\infty}\mu_1\left(\bigcup_{k=\ell}^{\infty}\bigsqcup_{n=N_k+1}^{\infty} E_{nk}\right)\leq\lim_{\ell\to\infty}\frac{1}{2^{\ell-1}}=0.$$ Now, Claim 2 implies that $\mu(S)=0$. Voi-effin'-là!

  • $\begingroup$ By assumption union of $E_nk$ for each k is $K_0$, so there is no need to invoke the third hyposethis. Hence there is no need to mention $f$ at all. $\endgroup$
    – Norbert
    Commented Jan 5, 2014 at 14:57
  • $\begingroup$ @Norbert I did so for completeness, because in the original problem I needed this result for, the $E_{nk}$'s were actually defined in terms of $f$ and their properties that I stated as assumptions here (in particular, their being of finite measure) had needed to be proven. $\endgroup$
    – triple_sec
    Commented Jan 5, 2014 at 18:23

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