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itemize 环境错位

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itemize 环境错位

我得到了一个意想不到的左对齐结果，我认为这是一个简单的 itemize 环境——以前从未遇到过这样的问题。以下是代码：

\begin{itemize} 
\item[(i)] Some text; \vspace{4pt}

\item[(ii)] Some text \vspace{4pt}

\item[(iii)]If $B_1, B_2 \in \mathfrak{B}(\mathbb{R})$ with $(B_1 \cap B_2) = \emptyset$, then\\ \vspace{3pt}
\hspace{2cm} $P_X(B_1 \cup B_2) = P_X(B_1) + P_X(B_2)$:\\ \vspace{3pt}
Since $B_1$ and $B_2$ are disjoint their pre-images in $\Omega$ will be disjoint and by the 
$\sigma$-additivity of $P$ on $\mathcal{A}$ we have that,\\ \vspace{3pt}
$P(X^{-1}(B_1 \cup B_2)) = P(X^{-1}(B_1)) + P(X^{-1}(B_2))\\ \vspace{3pt}
\hspace{2cm}  = P_X(B_1) + P_X(B_2)$. \\ 
\end{itemize}

问题出在第三项上。当以“Since”开头的文本（在换行符之后）换行时，它似乎与其余元素的开头左侧对齐。我不认为问题出在换行符上，因为之前这并没有给我带来问题。有什么想法吗？

答案1

也许你想使用显示数学。我认为没有理由使用这么多\vspace指令。让 TeX 进行对齐。

\documentclass{article}
\usepackage{amsmath,amssymb}

\begin{document}

\begin{itemize} 
\item[(i)] Some text;

\item[(ii)] Some text

\item[(iii)] If $B_1, B_2 \in \mathfrak{B}(\mathbb{R})$ with 
  $(B_1 \cap B_2) = \emptyset$, then
  \[
    P_X(B_1 \cup B_2) = P_X(B_1) + P_X(B_2)
  \]
  Since $B_1$ and $B_2$ are disjoint their pre-images in $\Omega$ will be disjoint 
  and by the $\sigma$-additivity of $P$ on $\mathcal{A}$ we have that
  \begin{align*}
    P(X^{-1}(B_1 \cup B_2)) &= P(X^{-1}(B_1)) + P(X^{-1}(B_2))\\
    &= P_X(B_1) + P_X(B_2).
  \end{align*}
\end{itemize}

\end{document}

在此处输入图片描述

您还可以使用enumitem以避免手动编号。

\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage{enumitem}

\begin{document}

\begin{enumerate}[label=(\roman*)]
\item Some text;

\item Some text

\item If $B_1, B_2 \in \mathfrak{B}(\mathbb{R})$ with 
  $(B_1 \cap B_2) = \emptyset$, then
  \[
    P_X(B_1 \cup B_2) = P_X(B_1) + P_X(B_2)
  \]
  Since $B_1$ and $B_2$ are disjoint their pre-images in $\Omega$ will be disjoint 
  and by the $\sigma$-additivity of $P$ on $\mathcal{A}$ we have that
  \begin{align*}
    P(X^{-1}(B_1 \cup B_2)) &= P(X^{-1}(B_1)) + P(X^{-1}(B_2))\\
    &= P_X(B_1) + P_X(B_2).
  \end{align*}
\end{enumerate}

\end{document}

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