Let $g : \mathbb R^2 \to \mathbb R$ be continuous and satisfy $g(x,y)\geqq 0$ for all $(x,y)\in \mathbb R^2$ and $\iint_{\mathbb R^2} g(x,y)dxdy=1.$
Define $g_t (x,y):=\dfrac{g\left(\frac{x}{t}, \frac{y}{t}\right)}{t^2}$ for $t>0.$
Then,
(i) Calculate $\iint_\mathbb {R^2} g_t (x,y) dx dy$.
(ii) Calculate $\displaystyle\lim_{t\to + 0} \iint_{D} g_t(x,y) dxdy$, where $D=\{ (x,y) \in \mathbb R^2 \mid x^2+y^2 \geqq 1\}$.
I did (i) and I found $\iint_\mathbb {R^2} g_t (x,y) dx dy=1.$
I haven't done (ii) yet.
Here is what I considered.
I think I can use (i).
$1=\int_{\mathbb R^2} g_t(x,y) dxdy=\int_{x^2+y^2 <1} g_t (x,y) dxdy+ \int_{D} g_t (x,y) dxdy$, thus $\int_{D} g_t (x,y) dxdy =1-\int_{x^2+y^2 <1} g_t (x,y) dxdy.$
If I use polar coordinates,
\begin{align} \int_{x^2+y^2 <1} g_t (x,y) dxdy &=\int_0^{2\pi} \int_0^1 g_t (r\cos \theta, r \sin \theta) r dr d\theta \\ &=\int_0^{2\pi} \int_0^1 \frac{r}{t^2} g\left(\frac{r \cos \theta}{t}, \frac{r \sin \theta}{t}\right) dr d\theta \\ \end{align}
And I want to calculate $\lim_{t \to +0}\int_0^{2\pi} \int_0^1 \frac{r}{t^2} g\left(\frac{r \cos \theta}{t}, \frac{r \sin \theta}{t}\right) dr d\theta$ but calculating this limit seems not to be easy.
I also tried with Taylor expansion of $g$ (at $(0,0)$) and this also seems not to work.
So, I'm having difficulty in calculating (ii). I'd like you to give me any help.
As regards (ii), after letting $X=x/t$ and $Y=y/t$, \begin{align} \int_{x^2+y^2<1} g_t (x,y) dxdy&=\int_{x^2+y^2 <1}\dfrac{g\left(\frac{x}{t}, \frac{y}{t}\right)}{t^2} dxdy=\int_{X^2+Y^2 <1/t^2}\dfrac{g\left(X, Y\right)}{t^2} t^2dXdY\\ &=\int_{X^2+Y^2 <1/t^2}g\left(X, Y\right)dXdY. \end{align} Hence \begin{align} \lim_{t \to 0^+}\int_{x^2+y^2<1} g_t (x,y) dxdy&= \lim_{t \to 0^+}\int_{X^2+Y^2 <1/t^2}g\left(X, Y\right)dXdY\\ &=\lim_{R \to +\infty}\int_{X^2+Y^2 <R^2}g\left(X, Y\right)dXdY= \iint_{\mathbb R^2} g(X,Y)dXdY=1 \end{align} and by (i) we may conclude that $$\lim_{t \to 0^+}\int_{D} g_t (x,y) dxdy=1-\lim_{t \to 0^+}\int_{x^2+y^2 <1} g_t (x,y) dxdy=0.$$