## Updated LaTex Javascript Converter

I have updated the LaTeX to WordPress Converter to change  to $l atex\displaystyle and$ respectively.

Note that $…$ is preferable over $$…$$.

Test code:

Input:

If $(X,\Sigma,\mu)$ is a measure space, $f$ is a non-negative measurable extended real-valued function, and $\epsilon>0$, then $\mu(\{x\in X: f(x)\geq\epsilon\})\leq\frac{1}{\epsilon}\int_X f\,d\mu.$

Define $s(x)=\begin{cases} \epsilon, &\text{if}\ f(x)\geq\epsilon\\ 0, &\text{if}\ f(x)<\epsilon. \end{cases}$
Then $0\leq s(x)\leq f(x)$. Thus $\int_X f(x)\,d\mu\geq\int_X s(x)\,d\mu=\epsilon\mu(\{x\in X: f(x)\geq\epsilon\})$. Dividing both sides by $\epsilon>0$ gives the result.

Output:

If $(X,\Sigma,\mu)$ is a measure space, $f$ is a non-negative measurable extended real-valued function, and $\epsilon>0$, then $\displaystyle \mu(\{x\in X: f(x)\geq\epsilon\})\leq\frac{1}{\epsilon}\int_X f\,d\mu.$

Define $\displaystyle s(x)=\begin{cases} \epsilon, &\text{if}\ f(x)\geq\epsilon\\ 0, &\text{if}\ f(x)<\epsilon. \end{cases}$
Then $0\leq s(x)\leq f(x)$. Thus $\int_X f(x)\,d\mu\geq\int_X s(x)\,d\mu=\epsilon\mu(\{x\in X: f(x)\geq\epsilon\})$. Dividing both sides by $\epsilon>0$ gives the result.