I understand that the expectation exist if and only if it is in $L_1$. i.e. $\int |x| dP < $ $\infty$. But our lecturer introduced us a new definition as follows:
For a discrete rv $X$ with support {$X_k$}, $E(X) $ is defined if and only if min{$ \sum_{k:x_k>0}$$ x_kP_x(x_k)$, -$ \sum_{k:x_k<0}$$ x_kP_x(x_k)$} $<$ $\infty$.
What I did not get is should it be max? Since the minimum of absolute value of positive and negative part would not guarantee the existence of expectation?
From the definition I was taught: $E(X)$ exists (assuming $X$ has countable support $\{x_i\}$) if and only if $\sum_i |x_i|P(x_i)$ converges. This is called absolute convergence (we don't need to take the absolute value of probabilities, since they are non-negative). Absolute convergence is equivalent to the convergence of both $\sum_{k:x_k>0} x_kP_x(x_k)$ and $\sum_{k:x_k<0} x_kP_x(x_k)$. So yes, you do need to take the maximum, not the minimum in the the way you've got it phrased to guarantee this.
To see why you might want absolute convergence in the definition, check out the Riemann rearrangement theorem.
Thanks Yacoub! I think I just figured out why my lecturer define it this way. (Meaning of non-existence of expectation?). The point is not make |x| to be integrable. What they are trying to avoiding is just not to have both + $\infty$ and - $\infty$ at the same time. :D Should I close this question?
It depends on the text. Some texts will say 'defined' or 'exists' to mean the same thing as integrable, i.e. excluding the case where $E[|X|] = \infty$. However, in your text, an infinite expectation is considered to be 'defined' though it doesn't 'exist' (as a (finite) real number).
Analogy in Calculus: Does $\lim_{x \to \infty} x $ 'exist' ?