This proposition was left for the reader to prove:

The following implications are valid if and only if the measure $\mu$ is complete:

a.) If $f$ is measurable and $f = g$ $\mu$-a.e., then $g$ is measurable.

b.) If $f_n$ is measurable for $n\in \mathbb{N}$ and $f_n\rightarrow f$ $\mu$-a.e., then $f$ is measurable.

I am not looking for an answer, just need some guidance or hint, I also fail to grasp the concept of the $\mu$ almost everywhere part. Any suggestions is greatly appreciated.

attempted proof of b.) $\Rightarrow$ We are given $f_n$ to be measurable for $n\in\mathbb{N}$, and $f_n\rightarrow f$ a.e. From proposition 2.7 we can let $$\hat{f} = \lim_{n\rightarrow \infty}\sup f_n$$ since $f_n$ is stated to be measurable, then $\hat{f}$ is also measurable. Also, since $f_n\rightarrow f$ a.e, we then have $\hat{f} = f$ a.e, so by part (a) $f$ is measurable.

$\Leftarrow$ suppose $\mu$ is not complete. Then there exists a measurable set $E$ such that $\mu(E) = 0$ and a set $F\subset E$ such that $F$ is not measurable. Then for (a), note that $1_F$ is not measurable and $1_F = 0$. Similarly, for (b) define $f_n = 0$ for all $n$ and $f = 1_F$


Note that one must be careful about what ae. means. When we say that $f=g$ ae., it means that there is a measurable set $E$ of measure zero such that $f(x)=g(x)$ for $x \notin E$.

Suppose $\mu$ is complete. Let $E$ be the exceptional set where $f(x) \ne g(x)$.

Suppose $A$ is measurable. Then $g^{-1}(A) = ( g^{-1}(A) \cap E) \cup ( g^{-1}(A) \cap E^c)$. The set $g^{-1}(A) \cap E$ is measurable since it is contained in $E$ which has measure zero. We have $g^{-1}(A) \cap E^c= f^{-1}(A) \cap E^c$, hence it is measurable and so $g^{-1}(A)$ is measurable, and so $g$ is measurable and so Part (a) holds.

Now suppose Part (a) holds. Let $N \subset E$, where $E$ has measure zero. Let $f=1_{E}$ and $g = 1_{N}$. Then $f=g$ ae. and so $g$ is measurable. Hence $g^{-1}(\{1\}) = N$ is measurable. Hence $\mu$ is complete.

Part (b) is similar. Note that if $h_n \to h$ with the $h_n$ measurable, then $h$ is measurable. So, this really can be reduced to Part (a) fairly easily.

  • $\begingroup$ I never heard of the term "exceptional set", what does that mean? $\endgroup$ – Wolfy Oct 30 '15 at 14:22
  • $\begingroup$ Just a descriptive term, it contains the elements that are exceptions to the rule $f(x)=g(x)$. $\endgroup$ – copper.hat Oct 30 '15 at 14:25
  • $\begingroup$ I am trying to understand why $g^{-1}(A)\cap E^c = f^{-1}(A) \cap E^c$ could you explain that part? $\endgroup$ – Wolfy Oct 30 '15 at 18:03
  • $\begingroup$ If $x \in g^{-1}(A) \cap E^c$ then $g(x) \in A$ and $x \in E^c$, so $f(x) = g(x) $ and we have $x \in f^{-1}(A) \cap E^c$. Similarly for the other direction. $\endgroup$ – copper.hat Oct 30 '15 at 18:04
  • $\begingroup$ I tried to prove part b for the if part, am I on the right track? $\endgroup$ – Wolfy Oct 30 '15 at 18:21

The $\mu$-a.e. condition says that $f=g$ everywhere except on some (possibly empty) set $S$, and $S$ is of measure zero according to measure $\mu$.

Couple this to the definition of measurable, and part A becomes easy.

Part B requires a bit more thought.

  • $\begingroup$ Been thinking about part a.) for almost an hour now still have no idea where to start, any other hints you can give me? $\endgroup$ – Wolfy Oct 30 '15 at 0:54

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