Prove that $F(x)=\int_{x_{1}^{0}}^{x_1}\cdots \int_{x_{n}^{0}}^{x_n} f(y) \, dy_1 \cdots dy_n$ is continuous

Question

Let $\Omega$ be a open subset of $\mathbb{R}^{n}$, $f \in L_{\operatorname{loc}}^{1}(\Omega)$ and $x^0=(x_{1}^{0},\dots,x_{n}^{0})$ an arbitrary point of $\Omega$. Define $$F(x)=\int_{x_{1}^{0}}^{x_1}\cdots \int_{x_{n}^{0}}^{x_n} f(y) \, dy_1 \cdots dy_n.$$

My question: How to prove that $F$ is a continuous function in a sufficiently small neighborhood of $x^0$?

I started trying to prove this statement in the case $\Omega=\mathbb{R}$. So, let $x_0 \in \mathbb{R}$ and $(x_n) \subset \mathbb{R}$ such that $x_n \rightarrow x_0$ in $\mathbb{R}$. Let $M>0$ such that $-M<x_n<M$ and $-M<x_1^0<M$ for all $n \in \mathbb{N}$. Then, for $x^0=x_{0}^{1}$ $$F(x_n)=\int_{x_1^0}^{x_n}f(y)dy=\int_{\mathbb{R}} 1_{(x_1^0,x_n)}(y)f(y)\,dy. \tag{*}$$ Then, define $g_n(y)=1_{(x_1^0,x_n)}(y)f(y)$. We have that $|g_n(y)|\leq 1_{(-M,M)}|f(y)|=g(y)$ and $g \in L^1(\mathbb{R})$. If we prove that $$1_{(x_1^0,x_n)} \rightarrow 1_{(x_1^0,x_0)} \hbox{ a.e. in } \mathbb{R},$$ the result follows from the Dominated Convergence Theorem. (This is also a point that I have not been able to prove.)

It's just weird to write (*) this because it can happen $x_{n_0-1}<x_1^0$ and $x_{n_0}>x_1^0$ for some $n_0 \in \mathbb{N}$.

PS: This question comes from Corollary 1, page 263 in Trèves book Topological Vector Spaces, Distributions and Kernels.

Answer 1

Let us first consider the case that $f$ is an $L^\infty$-function: We show that $F$ is a continuous function in a small neighborhood $B_\delta(x_0)$ such that $\overline{B_\delta(x_0)} \subset \Omega$. Let $u_0 \in B_\delta(x_0)$, then for any $u \in B_\delta(x_0)$ $$|F(u_0)-F(u)| \leq \|f\|_\infty \lambda([x_0,u_0] \Delta [x_0,u]), $$ where we write $[a,b] = [a^{(1)},b^{(1)}] \times \ldots \times [a^{(n)},b^{(n)}]$ and $A \Delta B = A \setminus B \cup B \setminus A$ for the symmetric difference.

In the last step, we have used that $$F(u)-F(u_0) = \int f (1_{[x_0,u]}-1_{[x_0,u_0]}) dy$$ and $$|1_{[x_0,u]}-1_{[x_0,u_0]}| = 1_{[x_0,u] \Delta [x_0,u_0]}.$$ The last estimate can be checked by considering all cases:

• If $x \in [x_0,u]$, but $x \notin [x_0,u_0]$, then the left term is $1$ and the right term is $1$ as well.
• The same holds if $x \notin [x_0,u]$, but $x \notin [x_0,u]$.
• If $x \in [x_0,u]$ and $x \in [x_0,u_0]$, then the first term is $|1-1|=0$ and the second as well.
• If $x \notin [x_0,u]$ and $x \notin [x_0,u_0]$, then both are zero.

Now note that $x \in [x_0,u_0] \Delta [x_0,u]$ implies that we have $x \in [x_0,u_0]$ or $x \in [x_0,u]$. We only consider the first case and then we must have $x \notin [x_0,u]$. This means that $x_i \in [x_0^{(i)},u_0^{(i)}]$ but $x_i \notin [x_0^{(i)},u^{(i)}]$ for some $i=1,\ldots,n$. (You may draw a picture of this set.) Thus, we must have $x_i \geq u^{(i)}$. Together with $x_i \leq u_0^{(i)}$ we get that $x_i$ varies in an interval of length $|u^{(i)}-u_0^{(i)}|$. Thus $$\lambda^n([x_0,u_0] \Delta [x_0,u]) = \lambda^n( \cup_{i=1}^n \{ | u_0^{(i)} \geq x_i \geq u^{(i)}\}) \\ \leq \sum_{i=1}^n \lambda^n\{ | u_0^{(i)} \geq x_i \geq u^{(i)}\}) \leq 2C \sum_{i=1}^n |u^{(i)}-u_0^{(i)}|,$$ where $$C= \max_{i=1,\ldots,n} \max \{ \prod_{j \neq i} |x_j-y_j| : x,y \in B_\delta(x_0)\}.$$ Thus, $F$ is even locally Lipschitz-continuous. (Note that the length of the $i$-th coordinte is small while the other coordinates are just bounded trivially.)

If $f$ is a locally $L^1$-function, we now that there exists an $K>0$ such that $$\int_{\{|f|>K\} \cap B_\delta(x_0)} |f| dy < \varepsilon/4.$$ Now apply the first step to the function $f 1_{\{|f|\leq K\} \cap B_\delta(x_0)}$. So both arguments together gives here $$|F(u_0)-F(u)| < \varepsilon/2 + 2K C \|u_0-u\|_1.$$ Thus, for $\|u_0-u\|_1 < \varepsilon/(4KC)$, we find that $|F(u_0)-F(u)| < \varepsilon$.