Definition of Scale median Lehmann, in Theory of Point Estimation p.212 (and also on p.169), defines scale median as the solution to: $${E(X)I(X\le c)} = {E(X)I(X\ge c)}$$
given $X$ is a positive random variable, and ${E(X)}< \infty$.
However, the expression does not make sense as it ${E(X)I(X\le c)}$ and ${E(X)I(X\ge c)}$, are two random variables whose support intersect only at c. Hence a solution can not exist in general. Am I missing something here? Should it be defined as the solution to $${E(XI(X\le c))} = {E(XI(X\ge c))},$$
instead?
 A: It is interesting that in Google scholar, no results are found for "scale median". Any way, the exact text from Lehmann and Casella's book is (p. 212, 2nd edition)

3.7 Let $X$ be a positive random variable. (a) If $EX < ∞$, then the value of $c$ that minimizes $E|X/c − 1|$ is a solution to $EXI (X ≤ c) = EXI (X ≥
 c)$, which is known as a scale median.

Note carefully that there are no parentheses separating $X$ and $I (X ≤ c)$ or enclosing them. This is confusing notation alright, and uncharacteristic of the specific book where in other cases of the use of the $E$ operator, parentheses, brackets and curly brackets are used (some authors do write $EX$ instead of $E(X)$, but not here).  
Taking clues from Did's answer and corrective comments in the math.SE linked thread, as well as here, it is indeed
$E\big[XI (X ≤ c)\big]$, since Did uses $E(X;X\leq c)$ which is equal to $E\big[XI (X ≤ c)\big]$, being the expected value restricted in the set $\{X\leq c\}$. The right-hand side integrals are also equal to $E\big[XI (X ≤ c)\big]$ and its counterpart, indicating just a typo from the part of the OP there.
