I have learned about measurability of functions, and a number of theorems depend on functions being measurable.
But I am not actually sure how "pathological" a function needs to get before becoming unmeasurable?
Are all real valued functions that people use in practical applications measurable?
Do you have to produce "cantor set-like" functions to get an unmeasurable function?
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The answer of course depends on what precisely do you mean by pathological. However, assuming that any function which is not computable in any reasonable way is pathological, and thus if a construction of a function relies crucially on some form of the axiom of choice, then that function is to be considered pathological, then you can rest assured all functions that arise in 'real life' are measurable. More precise mathematical formulations of this claim go back to the work of Solovay and Shelah, examining the required amount of choice to assure non-measurable functions exist. Some choice is certainly required.
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Define on $\mathbb{R}$ an equivalence relation by $x\sim y \Leftrightarrow y-x\in \mathbb{Q}$. Define a set $V$ by choosing exactly one representative from each equivalence class. One can prove that $V$ is a non-measurable set. Note though that to 'construct' $V$ we used the axiom of choice.
Thus as indicated in idm's answer, the function $1_v$ is non-measurable. You can't explicitly write down such a function though as you can't write down $V$. Non-measurable functions are strange and normally you do not encounter them. (See also José Carlos Santos' answer).
The set $V$ above is called a Vitali set. Restricting to the space $[0,1]$, it is a fun exercise to show that such a set is indeed non-measurable. (Where the measure is a translation invariant probability measure such that the measure of an interval is the usual measure).
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Firstly, to understand unmeasurable functions, you really need to understand unmeasurable sets. An unmeasurable function is one where the pre-image of some open set is unmeasurable. So an unmeasurable function defines at least one complementary pair of unmeasurable sets, maybe many more. Conversely, given an unmeasurable set, we can create its indicator function which will be unmeasurable.
Secondly, unmeasurable sets are much, much worse than Cantor sets. Cantor sets are measurable. They are produced as the limit of a countably infinite procedure involving open (or closed) sets as a starting point. As such they are Borel sets. Borel sets are all the sets you would come across starting out with intervals, halflines, single points, and then carrying out countably infinite many procedures of limits, unions, intersections, and so on. (If you could do uncountable infinitely many operations, you could end up with any set, since you could express any set as the union of uncountably many single points.) In the same way, by starting with well-behaved functions such as continuous functions, indicator functions of intervals, or similar, and carrying out countably many procedures of limits, maximums, sums, etc., you still have a Borel function.
To get something which isn't a Borel set, you need to use weird procedures, usually involving the axiom of choice, which aren't equivalent to countably many unions, intersections, etc. Bear in mind that countably many unions and intersections means that you can do infinite unions of infinite intersections of infinite unions of infinite intersections of infinite unions... There are lots of non-artificial constructions where Cantor sets are combined with other Cantor sets, arbitrarily many times. It's all still countable. So non-Borel sets are weird and hard to imagine. You need to have at least a little idea of set theory and what the AoC does/what constructions aren't possible without the AoC, just to start thinking about them.
Now, measurable sets are much more general than Borel sets. (And similarly, unmeasurable functions are much worsely behaved than non-Borel functions). A measurable set is potentially the symmetric difference of a Borel set, and an arbitrary set of measure zero. Here the arbitrary set is one of your strange Axiom of Choice things, just that it has to be a subset of a well-behaved Borel set of measure zero. So these sets are weird in lots of ways, just from the point of view of measure they are well-behaved, because the weirdness is confined to a set of zero measure.
So to make an unmeasurable set, you have to have a non-Borel construction, which doesn't have measure zero. The right way to understand them, is that you know that some must exist (if you believe in the Axiom of Choice), because not all sets/functions can be measurable. But it's quite hard to write down an example, and really hard to get a good intuition of what the thing you constructed is 'really like'.
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It should not be very pathological if you don't specify the $\sigma$-algebra to be the borel $\sigma$-algebra. Let $A=[0, 1]\subset \mathbb R$. Now define the $\sigma$-algebra $\mathcal F =\{\mathbb R, A, A^c, \emptyset \} $. Take any $B\notin \mathcal F$ for example $B=[2, 3]$ then $\mathbf{1}_B$ is not measurable.
They are so horrible that their existence depends upon the set theory that you are working with. That is, there are models of the Zermelo-Fraenkel set theory (without the axiom of choice) for which every function from $\mathbb R$ into $\mathbb R$ is measurable. Therefore, there is no risk of stumbling into a non-measurable function in ordinary mathematical practice.