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{\bf Question}

If $f$ is measurable prove that for all $a,b\epsilon{\bf R}\,\,\,
\{x|a\leq f(x)&lt;b\}$ is measurable.  Is the converse of this result
true?


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{\bf Answer}

$\{x|a\leq f(x)&lt;b\}=\{x|f(x)&lt;b\}\cap\{x|f(x)\geq a\}$

The converse is not true, for example, let ${\bf R^n_+}$ be the
half space $x_1&gt;0$.  Let $A$ be a non-measurable subset of  ${\bf
R^n_+}$.  Define $f:{\bf R^n} \rightarrow {\bf R^*}$ by

$f(x)=\left\{\begin{array}{ccl}0&amp;{\rm if}&amp;x\not\epsilon{\bf
R^n_+}\\ +\infty&amp;{\rm if}&amp;x\epsilon A\\ -\infty&amp;{\rm
if}&amp;x\epsilon{\bf R^n_+}-A\end{array}\right.$

${}$

Then for all $a,b\epsilon{\bf R}, \,\,\, \{x|a\leq f(x)&lt;b\}$ is
either $\phi$ or the complement of ${\bf R^n_+}$, both of which
are measurable.  However $\{x|f(x)&gt;0\}=A$ which is non-measurable.


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