It's a great question, but unfortunately, I'm going to give an advanced probability/a measure theory answer that answers for random variables in general (**OP's original title is actually about random variables. It was actually me, who edited it to discrete**). However, I really can't imagine an elementary answer that consists of something more than simply 'Because' (or in Tagalog/Filipino 'Kasi'. I guess this is more precise because Tagalog/Filipino also has a word 'dahil' for a different kind of 'because').

Elementary probability peeps (including OP, I guess) may ignore this, except possibly the 'last part'. Or maybe elementary probability peeps can try to skim this and try to get an idea or something instead of understanding the specific details:

Defining expectations of random variables (or Lebesgue integrals of measurable functions) is in steps defined in the standard machine:

- Expectation of indicator random variables
- Expectation of simple random variables
- Expectation of nonnegative random variables
- Expectation of general random variables.

After 1 and 2, we get to 3. Let $X$ be a random variable. Here, we can define $E[X^{+}]$ and $E[X^{-}]$ for the nonnegative random variables $X^{+}$ and $X^{-}$, where $X^{+}$ and $X^{-}$ are something like, resp, 'the nonnegative part of $X$' and the negative of the 'nonpositive part of $X$' s.t. $X = X^{+} - X^{-}$. Then, given $|X| = X^{+} + X^{-}$, we can define $E[|X|] := E[X^{+}] + E[X^{-}]$. (Wait I forgot. Maybe you don't define this. Maybe it's really just decomposed by linearity of expectation/integration...)

And then we get to 4:

$$E[X] := E[X^{+}] - E[X^{-}]$$

Now, this is defined for either

$X$ that satisfies $E[|X|] < \infty$ --> Here, we say that $X$ is Lebesgue integrable.

$X$ that satisfies $E[X^{+}] < \infty$ or $E[X^{-}] < \infty$ --> Here, we say that the Lebesgue integral of $X$ exists.

Here, Condition 1 implies ($E[X^{+}] < \infty$ and $E[X^{-}] < \infty$, which implies) Condition 2 but not conversely. I guess analogous to regular limits in elementary calculus it's like: $\lim x^2 = \infty$, so $\lim x^2$ 'exists' but is not 'existable'. Meanwhile, $\lim \frac1x = 0$, so $\lim \frac1x$ is 'existable'. So 'existable' is the same as the usual 'exists' in elementary calculus, but 'exists' here is like including $\pm \infty$.

**Finally, the aforementioned 'last part'**:

- Anyway, I believe most texts on measure theory or advanced probability will consider specifically the Lebesgue integrable ones and not all of the Lesbesgue-integral-exists ones. On wiki, it says 'It turns out that this definition gives the desirable properties of the integral.' I assume this statement refers to Lebesgue integrable.

So now this begs the question as to why those texts refer to Lebesgue integrable random variables/measurable functions specifically. Well, there's probably (lol) some properties that Lebesgue integrable's satisfy that not all of the Lesbesgue-integral-exists ones do, but I figure it's just to avoid $\pm \infty$

- AlohaSine mentioned about $\infty - \infty$ cases in h answer, but I think it doesn't apply to the Lesbesgue-integral-exists ones:

For $E[X^{+}] < \infty$ and $E[X^{-}] = \infty$, $E[X] = - \infty$

For $E[X^{+}] = \infty$ and $E[X^{-}] < \infty$, $E[X] = \infty$

Wait actually I just realised that even for elementary level probability, we use this kinds of definitions eg St. Petersburg paradox: $E[X]$ doesn't really 'exist' in the sense that $E[|X|] = \infty$, but we do have that $E[X]$ '$=\infty$' under the idea that $E[|X|] = \infty$ because $E[X^{+}] = \infty$ while $E[X^{-}] < \infty$ (specifically $E[X^{-}] =0 $) as I believe.

- there's a lot of stuff in comments and answers about rearrangement in re conditional convergence. Not so familiar about this apart from wiki, but I think this (explicitly) explains only the case of discrete random variables and not for random variables in general (whenever their expectations exist...or are 'existable')