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.

Advertisements

About mathtuition88

http://mathtuition88.com
This entry was posted in math and tagged , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s