I'm tackling the following question:
My approach to (a) was:
Let $x \in (0, \infty)$ and consider any sequence $x_j \to x$. We are asked to prove that $v(x_j) \to v(x)$ i.e. that
$$\int_\mathbb R \mathbb 1_{[0,x_j]}u \ d\lambda \to \int_X \mathbb 1_{[0,x]} u\ d\lambda$$
Define $f_j : \mathbb (0, \infty) \to \mathbb R$ by $f_j(x) = \mathbb 1_{[0,x_j]} u$. $[0,x_j]$ is in $\Sigma$ so $f_j$ is $\Sigma$-measurable.
Observe that $u$'s $\lambda$-summability and $\Sigma$-measurability implies that $|u|$ is also $\Sigma$-measurable, $\lambda$-summable. So we have that $|f_j| \leq |u|$ which is $\lambda$-summable, and that $f_j \to \mathbb 1_{[0,x]}u$ pointwise, meaning we can apply the Dominated Convergence Theorem to conclude the convergence.
However I now notice that I can't actually use DCT because $\mathbb 1_{[0,x_j]} u$ is not necessarily finite. I don't know where to go from here, and I also don't know why the answer to (b) would depend on which measure is in use. As far as I can tell, none of the standard convergence results (Fatou, MCT, DCT, etc.) are specific to the Lebesgue measure.
$|I_{[0,x_j]} u| \leq |u| $ so DCT can be applied. Finiteness of $u$ is not an issue since it is integrable, hence finite almost everywhere.
Part b) Let $\lambda(E)=1$ if $x_0 \in A$ ,$0$ otherwise. Then $v(x)=u(x_0)$ if $x \geq x_0$ and $0$ otherwise. This gives a counterexample for part b). [For a specific example take $x_0=1$ and $u(x)=I_{[0,2]}$].