proof using Brownian motion scaling property

Question

I am trying to prove $A_t^+$ has the same distribution as $tA_1^+$ using the scaling property, where $$ A_t^+ = \int_0^t\mathcal 1_{[0,\infty)}(W_s)\,ds $$ I tried to scale the Brownian motion but it cannot lead to the right conclusion. I wonder which step is wrong. Here is what I attempted. $$ \begin{aligned} tA_1^+ & = t\int_0^1\mathcal 1_{[0,\infty)}(W_s)\,ds \\ & = \sqrt{t}\int_0^1\mathcal 1_{[0,\infty)}(\sqrt{t} W_{s})\,ds \\ & = \frac{1}{\sqrt{t}}\int_0^1\mathcal 1_{[0,\infty)}(W_{ts}) \,d(ts) \\ & = \frac{1}{\sqrt{t}}\int_0^t\mathcal 1_{[0,\infty)}(W_u) \,du \\ & = \frac{1}{\sqrt{t}}A_t^+ \end{aligned} $$

Answer 1

You see, you let $\sqrt{t}$ enter the parenthesis of the indicator, which does not make sense. And you don't need that to have the indicator $1_{ \ge 0}(\sqrt{t}W_s)$ because it already is:
$$ 1_{ \ge 0}(W_s) = 1_{ \ge 0}(\sqrt{t}W_s) $$

Answer 2

For $t>0$. \begin{align} A_t^+ &= \int_0^t\mathbb{1}({W_s \geq 0)} ds\\ &=^{u = s/t} t\int_0^1\mathbb{1}({W_{ut} \geq 0})du\\ &= t\int_0^1{\mathbb{1}\left(\frac{W_{ut}}{\sqrt{t}}\geq 0\right)}du\\ &=^{(d)} t\int_0^1{\mathbb{1}(\tilde{W}_{u} \geq 0)}du\\ \end{align} In the last equality, we use the scaling property of the BM i.e. for $u\geq0$, $\tilde{W}_u = \frac{1}{\sqrt{t}}W_{ut}$ is again a BM.