Characterization of measurable functions Let $X$ be an uncountable set, let $\mathfrak{M}$ be the collection of all sets $E\subset X$ such that either $E$ or $E^c$ is at most countable, and define $\mu(E) = 0$ in the first case, $\mu(E) = 1$ in the second. Prove that $\mathfrak{M}$ is a $\sigma$-algebra in $X$ and that $\mu$ is a measure on $\mathfrak{M}$. Describe the corresponding measurable functions and
their integrals.
I showed that $\mathfrak{M}$ is $\sigma$-algebra and $\mu$ is a masure on $\mathfrak{M}$. But I have some problems with description of measurable functions.
Here's my thought: Let's consider measurable functions $f:X\to [-\infty,+\infty]$. If function $f$ is measurable then $f^{-1}(\{a\})\in \mathfrak{M}$ for any $a\in [-\infty,+\infty]$. Hence either $f^{-1}(\{a\})$ or $(f^{-1}(\{a\}))^c$ is at most countable.
What's next? I would be very thankful is somebody show to me a detailed proof of this. I thought on this problem about 1.5 day but no results.
 A: Suppose $f$ is measurable. Let $L_\alpha = f^{-1} ([-\infty, \alpha])$. Note that if $ \alpha \le \beta$ then
$L_\alpha \subset L_\beta$. Also note that either $L_\alpha$ or its complement is countable.
Note that $L_\infty=X$. 
Let $m = \inf \{ t |L_t^c \text{ is countable } \}$.
Note that $\mu L_t = 1$ for all $t > m$, and since $\mu X = 1$ we see that
$\mu L_m = \lim_n \mu L_{m+{1 \over n}} = 1$. In particular, $L_m$ is
uncountable.
If $t<m$ then $L_t$ is countable and $\mu L_t = 0$. Since
$\mu L_m = 1$, and $[-\infty,m] = [-\infty,m) \cup \{m\}$ we see that
$\mu \{ x | f(x) = m \} = 1$ or, equivalently, $\mu \{ x | f(x) \ne m \} = 0$.
We see that the measurable functions are those that are constant except at a (at most) countable number of points.
It should be clear that if $m $ is finite, then $\int m d \mu = m$ and
hence $\int f d \mu = m$.
We also see that a measurable $f$ is integrable iff $m$ is finite.
A: Hint: Show that every measurable function in this setting is $\mu$-almost surely constant, i.e. if $f \colon X \to [- \infty, \infty]$ is $\mathfrak M$-measurable, then there is a unique constant $c \in [- \infty, \infty]$ such that $\mu ( \{ x \in X \mid f(x) = c \}) = 1$.
Once you've verified this, the part regarding integrals is trivial: $\int f d \mu = c$, where $c$ is the unique constant from above.
