Let $g:X\times Y \rightarrow \mathbb{R}$, and define $f(x) = \int_{Y} g(x,y)\,dy$. Suppose that $g$ is of bounded variation, then can I say that $f(x)$ is also of bounded variation?
I have shown \begin{align*} TV(f,X) = & \sup_p \sum_{i=0}^{n_p-1} |f(x_{i+1})-f(x_i)|\\ \le & \sup_p \sum_{i=0}^{n_p-1} \int_{Y}|g(x_{i+1},y)-g(x_{i},y)|\,dy\\ \le &\text{I do not know how to proceed from here.} \end{align*}