stochastic fubini theorem

Question

Take two stochastic processes: $X_t=\int_0^tx_sdW_S$ and $Y_t=\int_0^t y_sdW_s$, where $W_s$ is a standard Brownian motion. Let $Z_t=X_tY_t$, and consider the following integral $\int_0^1 Z_t dt$. We can use the stochastic Fubini theorem here to show that $$ \int_0^1 Z_t dt = \int_0^1 \int_0^tdZ_s dt = \int_0^1 (1-t)dZ_s $$ But I can also rewrite $\int_0^1Z_tdt$ as $\int_0^1 \int_0^t X_t dY_s dt$. My question is can I now use the stochastic fubini theorem here to show $$ \int_0^1 Z_t dt = \int_0^1 X_t\int_0^t dY_s dt = \int_0^1\int_t^1X_s dsdY_t $$ I am confused because the integrability conditions for the Fubini theorem seem to be satisfied but $\int_t^1X_sds$ is no longer adapted to the filtration $\sigma(W_s:s\leq t)$. Is such a change in the order still permitted? If not, can you please tell me why it fails? If it is, how do I show $$ \int_0^1 (1-t)dZ_t = \int_0^1\int_t^1X_s dsdY_t $$