Continuity of $x\mapsto \lambda (X\cap ([0,x]\times[0,y]))$

Question

Denote by $\lambda$ the Lebesgue measure on $\mathbb{R}^2$ and let $X\subseteq [0,1]^2$ be a measurable bounded set

Is it true that the function $f:[0,1]^2 \to \mathbb{R}$ defined by $$ (x,y) \mapsto \lambda (X\cap ([0,x]\times[0,y])) $$ is continuous?

Edit: why these downvotes? Please let me know whether the question is not well posed and any detail is missing. The only non specified thing that I can see is that I am implicitely referring to the topology induced by the Euclidean norm. I know that the answer is positive in the case of $\mathbb{R}$, because continuity can be splitted in left and right continuity. But what about the case of higher dimension? A related case is provided in this MSE thread.

Answer 1

Set

$$f(x,y) := \lambda(X \cap ([0,x] \times [0,y]) = \int_X 1_{[0,x]}(r) 1_{[0,y]}(s)\, \lambda(dr,ds)$$

and fix $x_0,y_0 \in \mathbb{R}$. Then by the triangle inequality, $$\begin{align*} |f(x,y)-f(x_0,y)| &= \left| \int_X 1_{[0,x]}(r) 1_{[0,y]}(s) \, \lambda(dr,ds) - \int_X 1_{[0,x_0]}(r) 1_{[0,y]}(s) \, \lambda(dr,ds) \right| \\ &\leq \int_{[0,1] \times [0,1]} |1_{[0,x_0]}(r)-1_{[0,x]}(r)| 1_{[0,y]}(s) \, \lambda(dr,ds) \\ &\leq \int_{[0,1] \times [0,1]} |1_{[0,x_0]}(r)-1_{[0,x]}(r)| \, \lambda(dr,ds) \\ &\leq |x-x_0| \cdot 1 \end{align*}$$

for any $x,y \in \mathbb{R}$. An analogous calculation shows

$$|f(x_0,y)-f(x_0,y_0)| \leq |y-y_0|.$$

Hence,

$$\begin{align*} |f(x,y)-f(x_0,y_0)| &\leq |f(x,y)-f(x_0,y)| + |f(x_0,y)-f(x_0,y_0)| \\ &\leq |x-x_0| + |y-y_0| \end{align*}$$

which shows that $f$ is (Lipschitz) continuous.