Let $f$ be a compact support continuous function from $\mathbb R^n$ to $\mathbb R$ show that
$$ \lim_{k \to \infty} k^n \int_{(0,k^{-1})^n} f(x)dx=f(0)$$
Consider the change of variable $y=kx$ (scalar product by $k$) then $$k^n \int_{(0,k^{-1})^n} f(x)dx=\int_{(0,1)^n}f\left(\frac{y}{k} \right) dy=\int_{\mathbb R^n} f\left(\frac{y}{k} \right) \chi_{(0,1)^n}(y)dy$$ now let $g_k=f\left(\frac{y}{k} \right) \chi_{(0,1)^n}(y)$ is easy to see that $g_k \to f(0) \chi_{(0,1)^n}(y)$ and that $|g_k| \leq M \chi_{(0,1)^n}(y)$ (the constant $M$ is obtained because $f$ has compact support) and the function $M \chi_{(0,1)^n}(y)$ is in $L^1(\mathbb{R}^n)$ then by the dominated convergence theorem i conclude that $$\lim_{k \to \infty} \int_{\mathbb R^n} f\left(\frac{y}{k} \right) \chi_{(0,1)^n}(y)dy=f(0)$$ hence $$ \lim_{k \to \infty} k^n \int_{(0,k^{-1})^n} f(x)dx=f(0)$$ these is correct? any suggestion or hint i will be very grateful