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Q1) Let $X$ a r.v. Do we always have that $\mathbb E[X\boldsymbol 1_{[a,b]}]<\infty $ ?

Q2) Let $f:[a,b]\longrightarrow \mathbb R$ a function (not necessarily continuous). Does $\sup_{[a,b]}f<\infty $ ?

My attempts

1) Since $X(\omega )\in \mathbb R$ for all $\omega $, I woud say yes.

2) Since $f(x)\in \mathbb R$ for all $x\in [a,b]$ I would say yes.

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    About Q2. 'Not necessarily continuous' is an important remark. Take any function on (a, b) that has infinite supremum, such as tan. Now to make it into a function on all of [a, b] you just have to 'invent' values $f(a)$ and $f(b)$. When you require continuity this is impossible, when you don't require continuity this is very easy and you can do whatever you want... – 2017-01-16
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    For the first: I would start with the inequality $\lvert X\rvert \mathbf{1}_{[a,b]} \leq \max(\lvert a\rvert, \lvert b\rvert) \mathbf{1}_{[a,b]}$. – 2017-01-16
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    @Clement C., wait, why would that be? Take any continuous $X\colon [0,1]\to\Bbb R$ with large enough maximal value. – 2017-01-16
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    I may have mistaken the notation -- I am assuming $X \mathbb{1}_{[a,b]}$ refers to to $X \mathbb{1}_{[a,b]}(X)$. – 2017-01-16
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    It says that $X$ is any random variable. – 2017-01-16
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    I read that. I am then wondering how you interpret the question: what is the expectation taken over? (i.e., what is for you the probability space, and what is the argument of the indicator function?). Again, I assumed the above, but am wondering what would then be a standard parsing of the OP's question. – 2017-01-16
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    Isn't it just $X1_{[a,b]}(\omega) = X(\omega)1_{[a,b]}(\omega)$? I mean, you can't really use $X$ as random variable (as in Borel measurable function) and as argument of some function. – 2017-01-16
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    Another thing you might do is look up the "Cauchy distribution". – 2017-01-16
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    @Ennar Note that a Borel measurable function (like $\mathbb{1}$) acting on a Borel (alternately Lebesgue) measurable function ($X$) yields an overall Borel (or Lebesgue) measurable, and thus $\mathbb{1}_{[a.b]}(X)$ makes perfect sense as a random variable whenever $[a,b]$ has a meaning in $X$'s co-domain. Further, probabilistically, one does not engage with the sample space, – 2017-01-16
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    and random variables should be agnostic to the exact particulars of the sample space, especially things like whether $\Omega$ admits an ordering or not, which is required to grant meaning to $[a,b]$ within $\Omega$. Thirdly, one should note that the indicator notation is rampant in undergrad probability, where one usually has little clue of the notion of a measurability and associated complications. All of which is to argue that the interpretation should be $\mathbb{1}_{[a,b]}(X)$, and not as you determine. – 2017-01-16
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    @stochasticboy321, well, it is possible, and needs confirmation, then. But, personally, I'd find it terrible notation for composition. – 2017-01-16

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Q1) $\Bbb E[X1_{[a,b]}] = \int_a^b XdP$ by definition. $X(\omega)\in\Bbb R$ has nothing to do with the integral being finite. For example, $$\int_0^1\frac{dx}x = \lim_{t\to 0}\int_t^1\frac{dx}x = \lim_{t\to 0}(\left.\ln x\right|_t^1) = -\lim_{t\to 0}\ln t=+\infty$$ Can you use this to find appropriate probability space and appropriate random variable $X$ with no finite expected value?

Q2) Consider function $$f(x) =\left\{ \begin{array}{c l} \frac{1}{2x-(a+b)}&\!\!\!\!,\ x\neq \frac{a+b}2\\ 0&\!\!\!\!,\ x= \frac{a+b}2 \end{array} \right.$$