Straight to the point
Suppose $\mu$ is a measure on a measure space $Q$,$X$ is a topological vector space on which $X^*$ separates points, and $f$ is a function from $Q$ into $X$ such that the scalar function $\Lambda f$ are integrable with respct to $\mu$, for every $\Lambda \in X^*$; If there exists a vector $y$ such that $$\Lambda y = \int_Q (\Lambda f) d\mu$$ for every $\Lambda \in X^*$ we define $$ \int_Q f d\mu = y $$
Now assuming such $y$ exists it is claimed that $y$ is unique due to the separation of $X^*$. Why is that?
Here is my attempt
If $X^*$ separates points then for every $0 \neq y \in X$ there's at least one $X^*$ such that $\Lambda y \neq 0$. By contradiction if uniqueness wouldn't be true we could get $\Lambda y_1 = \Lambda y_2$ with $y_1 \neq y_2$ but this would imply $\Lambda(y_1 - y_2) = 0$ for every $\Lambda$, but this is false because of the separation property.
Is my proof correct? is basically a straight forward application of the separation property.