Continuity of a function defined by the Lebesgue measure of a closed ball My task is to prove that the function given by $f(x) = m(B_{x}(\alpha))$ is a continuous function, where $m(\cdot)$ denotes the Lebesgue measure and $B_{x}(\alpha)$ is a closed ball of radius $x$ centered at $\alpha$, i.e.
$$ B_x(\alpha) = \{ y \in \mathbb{R}^d \colon \Vert y-\alpha \Vert \leq x\} $$
The definition of continuity that I am using is as follows: A function $f(x)$ is continuous if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $\vert x - a \vert < \delta$ implies $\vert f(x) - f(a) \vert < \varepsilon$.
My main point of struggle is determining a $\delta$ that makes this all work nicely. Here is what I have currently:
Proof: Let $ \vert x - a \vert < \delta - a$. Without loss of generality, assume that $a < x$, and so $x - a < \delta - a$. Additionally, we may assume that $\alpha$ above equals $0$ without loss of generality. Consider any element $u \in \mathbb{R}^d$ such that $0 < \Vert u \Vert - a \leq x-a < \delta - a$. This implies that $a < \Vert u \Vert \leq x < \delta$. We may write the set of all possible $u$ as
$$ B_x(0) \backslash B_a(0) = \{ u \in \mathbb{R}^d \colon a < \Vert u \Vert \leq x \} $$
By the properties of Lebesgue measure, we have
$$m(B_x(0) \backslash B_a(0)) = m(B_x(0)) - m(B_a(0)) =  \vert m(B_x(0)) - m(B_a(0)) \vert$$
which is precisely $\vert f(x) - f(a) \vert$. But $B_x(0) \backslash B_a(0) \subseteq B_\delta(0)$ and so
$$m(B_x(0) \backslash B_a(0)) < m(B_\delta(0)) = \frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2} + 1)} \delta^d$$
Hence we have$\vert f(x) - f(a) \vert < \frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2} + 1)} \delta^d$.
If I have done everything else correctly, then this is where I am stuck. I am no expert when it comes to spheres of dimension $d$ but I think I have written the "volume" (or better yet, measure) of $B_\delta(0)$ correctly in the above semi-proof. Any direction in choosing $\delta$ to get the desired "less than $\varepsilon$" condition would be greatly appreciated.
 A: We have $$f(x) = m(B_x(\alpha)) = m(B_x(0)) = x^d m(B_1(0)).$$
Hence $f$ is a polynomial.
A: Suppose $R= [l_1,u_1] \times [l_d,u_d]$ is a rectangle, we have $\operatorname{vol} R = (u_1-l_1) \cdots (u_d-l_d)$.
Let $\Lambda = \operatorname{diag} (\lambda_1,...,\lambda_d)$, where each $\lambda_k > 0$. Consider the set $\Lambda R = \{ \Lambda x | x \in R \} = [\lambda_1 l_1, \lambda_1 u_1] \times [\lambda_d l_d, \lambda_d u_d]$, we see that
$\operatorname{vol} (\Lambda R) = (\lambda_1 \cdots \lambda_d) \operatorname{vol} R$.
Using the Lebesgue outer measure, we see that for any measurable set $A$ we have (I have omitted some details)
$m A = \inf\{ \sum_k \operatorname{vol}(R_k) \mid A \subset \cup_k R_k \}$.
Let ${\cal C} (A) = \{ \{ R_k\} \mid A \subset \cup_k R_k \}$, the collection of all countable covers of $A$ by rectangles.
It is straightforward to see that ${\cal C} (\Lambda A) = \{ \{ \Lambda R_k  \}_k \mid \{ R_k  \}_k \in {\cal C} (A) \}$ and hence
$m(\Lambda A) = \inf\{ \sum_k \operatorname{vol}(\Lambda R_k) \mid A \subset \cup_k R_k \} = (\lambda_1 \cdots \lambda_d) mA$.
Note that by translation invariance, $f(x) = m B_x(\alpha) = B_x(0)$, and since
$B_x(0) = \operatorname{diag} (x,...,x)B_1(0)$, we see that
$f(x) = x^d mB_1(0)$ from which continuity follows.
Note:
The above is a special case of a more general result which is that for any measurable $A$ and linear transformation $L$ we have
$m(LA) = |\det L| m(A)$.
Alternative proof:
Note that for any sequence $x_n \to x$, the indicator functions converge ae., that is
$1_{B_{x_n}(\alpha)} \to 1_{B_{x}(\alpha)}$ (if $x $ is on the boundary it need not converge).
Hence the dominated convergece theorem gives $m(B_{x_n}(\alpha)) = \int 1_{B_{x_n}(\alpha)}  \to \int 1_{B_{x}(\alpha)} = m(B_{x}(\alpha)) $.
