Image of a measurable function is not measurable

Question

I am trying to find an example that shows the image of a measurable function is not measurable.

Here is what I found: Let $X=\{0,1\}$ with $\sigma$-algebra $\{\emptyset, X\}$. Let $f:X \rightarrow X$ and $f(x) = 0$. Then $f(X) = \{0\}$ is not measurable.

But how do we know that $f$ is measurable in the first place? Obviously, we have $f^{-1}(\{\}) = \{\}$. However, how can we say that $f^{-1}(X)=f^{-1}(\{0,1\}) = X = \{0,1\}$? I mean $f$ is not surjective; it is impossible that $f(x) = 1$.

Thanks.

Answer 1

A function is measurable iff the preimage of any measurable set is measurable. That means that in this case, you only need to check that $f^{-1}(X)$ and $f^{-1}(\varnothing)$ are measurable (as $X$ and $\varnothing$ are the only sets which are measurable in this case). Since $f^{-1}(X) = X$ and $f^{-1}(\varnothing) = \varnothing$, we see that $f$ is indeed measurable. This is one of many examples which demonstrates how the preimage is a generally nicer operation than the forward image, especially for non-injective functions.

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Edit: to elaborate on why $f^{-1}(X) = X$, firstly note that $f^{-1}(X) \subset X$ is trivial. For the reverse inclusion, begin with $x \in X$. In order for $x \in f^{-1}(X)$ to hold, it is necessary that $f(x) \in X$. This is guaranteed by the fact that the codomain of $f$ is $X$.

This is the definition of the preimage: $a \in f^{-1}(A) \iff f(a) \in A$.

Answer 2

Remember that $f^{-1}(S)$ just means "all the points $x \in X$ such that $f(x) \in S$". You know that $f(0) \in X$ and $f(1) \in X$, so $0,1 \in f^{-1}(X)$. It's true that $f^{-1}(\{1\})=\emptyset$, but that isn't important.