+ \infty.$$

\[Proof of Theorem \[main\] using Proposition \[mainest1\] and Proposition \[mainest2\]\] The two propositions tell us that there is a point $x \in E_2$ so that

$$\label{L1close} \| d^x_*(\mu_1) - d^x_*({\mu_{1, good}}) \|_{L^1} < 1/1000, \textrm{ and}$$

$$\label{L2bound} \| d^x_*({\mu_{1, good}}) \|_{L^2} < + \infty.$$

Since $d^x_*(\mu_1)$ is a probability measure, (\[L1close\]) guarantees that

$$\int |d^x_*({\mu_{1, good}})| \ge 1 - \frac{1}{1000}.$$

Note that the support of $d^x_*(\mu_1)$ is contained in $\Delta_x(E)$. Therefore

$$\int_{\Delta_x(E)} | d^x_* {\mu_{1, good}}| = \int |d^x_* ({\mu_{1, good}})| - \int_{\Delta_x(E)^c} |d^x_*({\mu_{1, good}})|$$

$$\ge 1 - \frac{1}{1000} - \int |d^x_*(\mu_1) - d^x_*({\mu_{1, good}})| \ge 1 - \frac{2}{1000}.$$

But on the other hand,

$$\label{yes} \int_{\Delta_x(E)} | d^x_* {\mu_{1, good}}| \le | \Delta_x(E)|^{1/2} \left( \int |d^x_* {\mu_{1, good}}|^2 \right)^{1/2}.$$

Since (\[L2bound\]) tells us that $\int |d^x_* {\mu_{1, good}}|^2$ is finite, it follows that $|\Delta_x(E)|$ is positive.

To end this section, let us make some comments about the proofs of Proposition \[mainest1\] and Proposition \[mainest2\]. To prove Proposition \[mainest1\], the first observation is that if $x$ is far from $T$, then removing $M_T \mu_1$ from $\mu_1$ has a negligible effect on the pushforward measure $d^x_* (\mu_1)$. So the difference between $d^x_* \mu_1$ and $d^x_* {\mu_{1, good}}$ only comes from 