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Let $(\Omega, \Sigma)$ be a measurable space. Is the space of bounded measurable functions $B_b(\Sigma)$ equipped with the supremum norm a Banach space, i.e. complete?

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  • $\begingroup$ Yes. First: show that the pointwise limit of a Cauchy sequence exists, second: show that the pointwise limit is bounded, third: show that the pointwise limit is measurable. Fourth: show that the pointwise limit is a uniform limit. Which part is causing trouble? $\endgroup$ – t.b. Apr 3 '12 at 7:36
  • $\begingroup$ Look here: math.stackexchange.com/questions/71121/… $\endgroup$ – Rudy the Reindeer Apr 3 '12 at 7:36
  • $\begingroup$ (just replace continuous with measurable) $\endgroup$ – Rudy the Reindeer Apr 3 '12 at 7:37
  • $\begingroup$ @tb: did you mean "Show that uniform limit (when exists) is a pointwise limit?" $\endgroup$ – Ilya Apr 3 '12 at 9:23
  • $\begingroup$ @Ilya: No. See the answer below. $\endgroup$ – t.b. Apr 3 '12 at 11:03
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Since my comment seems to have caused some confusion, here's the argument (the answer is yes, of course):

Let $(f_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $B_{b}(\Omega,\Sigma)$, that is to say:

For every $\varepsilon \gt 0$ there exists $N(\varepsilon)$ such that for all $m,n \geq N$ we have $\|f_n - f_m\|= \sup_{x \in \Omega}{|f_n(x) - f_m(x)|} \lt \varepsilon$.

  1. For each $x \in \Omega$ we have $|f_n(x) - f_m(x)| \leq \|f_n - f_m\|$. It follows that $(f_n(x))_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$. Therefore $f(x) = \lim_{n \to \infty}{f_n(x)}$ exists (i.e. there is a function $f: \Omega \to \mathbb{R}$ such that $f_n \to f$ pointwise).

  2. The pointwise limit of a sequence of measurable functions is measurable: For all $a \in \mathbb{R}$ we have $$ \begin{align*} \{x \in \Omega\,:\,f(x) \leq a\} & = \left\{ x \in \Omega \,:\, (\forall k \in \mathbb{N}) \, (\exists m \in \mathbb{N}) \, (\forall n \geq m)\quad f_n(x) \leq a+ \frac{1}{k} \right\} \\ & = \bigcap_{k = 1}^\infty \bigcup_{m = 1}^\infty \bigcap_{n = m}^\infty \left\{x \in \Omega \,:\, f_n(x) \leq a + \frac{1}{k}\right\} \in \Sigma. \end{align*} $$ It follows that $f$ is measurable.

  3. For $m,n \geq N(\varepsilon)$ we have for all $x \in \Omega$ that $|f_n(x) - f_m(x)| \lt \varepsilon$. Letting $n \to \infty$ we see that $$ |f(x) - f_m(x)| \leq \varepsilon \quad\text{ for all } x \in \Omega\text{ whenever } m \geq N(\varepsilon). $$ It follows that

    • for all $x \in \Omega$ we have $|f(x)| \leq |f_m(x)| + \varepsilon \leq \|f_m\| + \varepsilon$, so that $f$ is bounded and by point 2 we already know that $f$ is measurable, so $f \in B_b(\Omega,\Sigma)$.

    • If $m \geq N(\varepsilon)$ we have $\|f - f_m\| \leq \varepsilon$ so that $f$ is the limit of $(f_n)$ in $B_b(\Omega,\Sigma)$.

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  • $\begingroup$ Great answer, I don't understand why it hasn't been accepted! $\endgroup$ – Chaos Mar 5 '20 at 13:22

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