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I was reading this link with the same question, but I couldn't quite understand what is happening. I will try to put my thoughts here.

Suppose $X$ is the random variable for a coin toss. Then $\Omega = \left\{1,2 \right\}$ and $\mathcal{F}=\left\{ \phi, \left\{1\right\}, \left\{2 \right\} , \Omega \right\}$.

If we define $X: \Omega \rightarrow \mathbb{R}$, we can recover the probabilities for all events in the event space $\mathcal{F}$.

However, if we define $X: \mathcal{F} \rightarrow \mathbb{R}$, then we would also have to map $\phi$ and $\Omega$ to $\mathbb{R}$. While for the first case we can simply assign these probabilities as $0$ and $1$, I don't understand why we can't do the same for this case.

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    I think you must be fairly close to understanding the purpose of assigning real values to items in $\Omega$. Trying to map the family of subsets $\mathcal{F}$ invites inconsistent and other meaningless valuations. We know that an outcome of a coin flip should be binary (heads or tails), and the mapping $X:\Omega \to \mathbb{R}$ models that. – 2017-01-18

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A random variable does not need to take a definite value in $\mathbb{R}$ on a given event, so it doesn't make sense to define it as a map from $\mathcal{F}\rightarrow \mathbb{R}.$

In your example, the events $X=H$ and $X=T$ are synonymous with the outcomes. Consider instead two coin tosses. Then "the first coin flip is heads" is an event, but your random variable can be the value of the 2nd coinflip. It wouldn't make sense to assign it a value in $\mathbb{R}$.

Your suggestion of making $X(\Omega) = 1$ and $X(\emptyset) = 0$ in the case of one coinflip likewise does not make sense. What would that mean? The way I would roughly interpret it is "if X is known to be either heads or tails, then $X=1$" and "if nothing is known about $X$ then $X=0"$. That doesn't make any sense.

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    Just to clear up my remaining doubts, what do events $\phi$ and $\Omega$ mean? – 2017-01-18
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    Also, I want know, can we assign probabilities $P(\phi)$ and $P(\Omega)$ as $0$ and $1$ respectively if defined as the second case? Why/why not? – 2017-01-18
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    @radm94 The event $\phi$ is just an empty event with probability zero. As an event, it's mostly there for formal convenience, but when you interpret events as information about the outcome, it means 'no information'. The event $\Omega$ means 'any outcome happened', or in terms of information 'we have all the information about the outcome'. In general the event $F\in \mathcal{F}$ means "one of the outcomes in $F$ happened." – 2017-01-18
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    @radm94 Yes, it makes sense to say, and is always true that $P(\emptyset) = 0$ and $P(\Omega) = 1.$ Unlike random variables, probability maps from events, not outcomes. Likewise for the other two events in $\mathcal{F}$, $P(\{X=1 \}) = P(\{X=2\})=1/2$. Often it seems like they're mapping from outcomes, cause we are lazy and write $P(1) = P(2) = 1/2$ but we really mean the probability of the events $\{1\}$ and $\{2\}.$ It's confusing cause some events correspond to a single outcome, but not all events, as in my example with the two coinflips. – 2017-01-18