math blocks in standard obsidian format doesn't render

[Masacroso]

For some reason my maths blocks doesn't render, however it does in obsidian. The standard way to write math blocks in obsidian is something like

paragraph
$$
maths
$$
paragraph

However I see that with this format the math blocks doesn't render in github either, so I change to the format

paragraph

$$maths$$

paragraphs

Now it renders correctly. However all my notes are written in the first style, there is a way to make that your template supports the use of the first format for math blocks instead of the second?

[jobindjohn]

I not sure why it doesn't work for you. It has been working for me. See the example here: https://jobindjohn.github.io/obsidian-publish-mkdocs/Features/LaTeX%20Math%20Support/

Can you verify if it's not rendering for you after it has been built

[Masacroso]

I not sure why it doesn't work for you. It has been working for me. See the example here: https://jobindjohn.github.io/obsidian-publish-mkdocs/Features/LaTeX%20Math%20Support/

Can you verify if it's not rendering for you after it has been built

Im not sure what you mean by "after build". The building is automatic in github after I push the files, I could see the files but they doesn't render block maths, just inline ones. I can give you a piece of text so you can debug by yourself what is going on (if you want/have time, of course):

## 2B.9

Because $\lambda ([0,1])=1$ then there is some non-measurable $A\subset [0,1]$. Then define $f:=\chi _A-\chi _{[0,1]\setminus A}$, then $f^{-1}(1)=A$, so $f$ is not measurable, however $|f|=\chi _{[0,1]}$.

---

## 2B.10

Let $A_{n,k}$ the set of numbers in $(0,1)$ with exactly $n$ fives in it firsts $k$ digits in it decimal expansion. It is easy to check that each $A_{n,k}$ is Borel and $A_n:=\bigcap_{k\geqslant 0}A_{n,k}$ is the set of numbers in $(0,1)$ with exactly $n$ fives, so is also Borel.

By last $A:=\bigcup_{n\geqslant 0}A_n$ is the set of numbers in $(0,1)$ with finitely many fives in it decimal expansion, and so it is also Borel, and clearly $A^\complement$ is also Borel.∎

---

## 2B.18

If $f$ is differentiable then it is continuous, so it is Borel measurable. Now note that
$$
f'=\lim_{n \to \infty }n(f(\cdot +1/n)-f)
$$
and every $n(f(\cdot +1/n)-f)$ is also Borel measurable.

---

## 2B.20

Note that $f^g=\exp\left(g\cdot \ln f\right)$, and so everything reduces to see that $\ln f$ and $e^h$ are measurable when $f$ and $h$ are.

---

## 2C.1

Because $\mu (X)\in[0,1)$, but this would imply that there are uncountable measurable subsets $E_t \subsetneq X$ such that $\mu (X)<\mu (E_t)$, what cannot be possible.

---

## 2C.3

Let $\mu (\{k\}):=2^{-k-1}$, now note that $\sum_{k\geqslant 1}2^{-k}=1$ and that each number $E\subset \Bbb N_{\ge 0}$ defines the binary expansion of a number in $(0,1]$. By last set $\mu (\emptyset ):=0$ to complete the measure.

---

## 2C.8

Let $X:=(0,1)$ and $\mathcal{A}:=\{(0,a/2),(a/4,a)\}$ for some $a\in(0,1)$. Then note that
$$
\sigma (\mathcal{A})=\mathcal{A}\cup \{(0,a/4),(a/4,a/2),(a/2,a),(a,1),\ldots ,(0,1),\emptyset \}
$$
Then fixing the values of $\mu$ and $\nu$ in $\mathcal{A}$ and $X$ we have a degree of freedom to choose the values of $(a/4,a/2)$ and $(a,1)$. ∎

---

## 2C.12

Because every singleton is measurable then there is some $f:X\to [0,\infty ]$ such that $\mu (\{x\})=f(x)$, and so it is easy to check that each measure in $(X,\mathcal{S})$ have the form
$$
\mu (E)=\alpha I(E)+\sup\left\{\sum_{x\in  D}f(x):D\subset E\,\land\, \#D\leqslant \aleph _0\right\}
$$
for arbitrary $\alpha \in \Bbb R$, and where
$$
I(E):=\begin{cases}
0, & \#E\leqslant \aleph _0\\
1, &  \text{ otherwise }
\end{cases}
$$
for each $E\in \mathcal{S}$.

---

## 2D.1

In the next $a_k$ count the number of lists of length $k$ of decimal digits that doesn’t contain a sub-list of consecutive 100 fours:
$$
\begin{align*}
&a_{k+1}=9a_k-8a_{k-100}[k\geqslant 100]-[k=99],\quad a_0=1,\quad A(x):=\sum_{k\geqslant 0}a_kx^k\\
&\therefore\, \sum_{k\geqslant 0}a_{k+1}x^k=9A(x)-8\sum_{k\geqslant 100}a_{k-100}x^k-x^{99}\\
&\iff A(x)-1=9xA(x)-8x^{101}A(x)-x^{100}\\
&\iff A(x)=\frac{1-x^{100}}{1-9x+8x^{101}}
\end{align*}
$$
Now let $B_k$ for the set of numbers in $(0,1)$ such that the first sub-list of 100 consecutive 4s start at the $k+1$ digit of it decimal expansion, and so the decimal expansion of a number in $B_k$ have the form
$$
\underbrace{0,d_1d_2\ldots d_{k-1}}_{\text{ no sublists of 100 consecutive 4s }}\overbrace{d_k}^{\text{ any digit but 4 }}
\underbrace{d_{k+1}\ldots d_{k+100}}_{\text{ 4s }}\overbrace{d_{k+101}\ldots }^{\text{ anything }}
$$
Thus it is easy to check that
$$
\lambda (B_k)=\frac{8a_{k-1}[k\geqslant 1]+[k=0]}{10^{k+100}}
$$
(a) The set that we are searching for is $\bigcup_{k\geqslant 0}B_k$, so it is clearly Borel.

(b) Because $B_k\cap B_j=\emptyset$ whenever $k \neq  j$ we find that
$$
\begin{align*}
\lambda \left(\bigcup_{k\geqslant 0}B_k\right)&=\frac1{10^{100}}+\frac{8}{10^{101}}\sum_{k\geqslant 1}\frac{a_{k-1}}{10^{k-1}}\\
&=\frac1{10^{100}}+\frac{8}{10^{101}}\left[\sum_{k\geqslant 0}a_kx^k\right]_{x=1/10}\\
&=\frac1{10^{100}}+\frac{8}{10^{101}}A(1/10)\\
&\approx 9\cdot 10^{-100}
\end{align*}
$$
---

## 2D.23

Note that $\Bbb R /\Bbb Q$ is uncountable, and so there is a bijection $g:\Bbb R /\Bbb Q \to \Bbb R$. Now let $f(x):=g([x])$.∎

---

## 2E.8

It is enough to note that, because $\sum_{k\geqslant }2^{-k}=2$ then for any chosen $\varepsilon >0$ there is some $N\in \Bbb N$ such that $0\leqslant 2-\sum_{k=0}^{N}2^{-k}<\varepsilon$, so choosing the finite set defined by $N$ we have finished.

[jobindjohn]

Interesting. In your example, you get an error only on 2D.1. Perhaps it has something to do with \begin{align*} ?

[Masacroso]

Interesting. In your example, you get an error only on 2D.1. Perhaps it has something to do with \begin{align*} ?

I get errors in the whole document, not just in 2D.1, in many places there is no \begin{align* }.

[mendelabramzon]

@Masacroso did you find a way to fix it (not manually I mean)? I also have structure
$$
\mathbb{R}
$$
and not $$\mathbb{R}$$ And it does not render