Having problems with align*

You probably want all columns aligned to the left, hence two additional &s:

\documentclass{article}

\usepackage{amsmath}

\begin{document}
\begin{align*}
    \label{fourth term 0}
 &   \lim_{\Delta t \to 0}\Bigg|\sum_{j=0}^{n-1}f_{tx}\big(t_j,X(t_j)\big)(t_{j+1} - t_j)\big(X(t_{j+1})-X(t_j)\big)\Bigg|\\
    &\leq \lim_{\Delta t \to 0}\Bigg|\sum_{j=0}^{n-1}f_{tx}\big(t_j,X(t_j)\big)(t_{j+1} - t_j)\big(X(t_{j+1})-X(t_j)\big)\Bigg|\\
    &\leq \lim_{\Delta t \to 0}\max_{0 \leq k \leq n-1}|X(t_{k+1}-X(t_k)|\cdot \lim_{\Delta t \to 0}\sum_{j=0}^{n-1}|f_{tx}\big(t_j,W(t_j)\big)|(t_{j+1}-t_j)\\
  &  = 0.\int_{0}^{T}\big|f_{tx}\big(t,W(t)\big)\big|dt = 0.%\numberthis
\end{align*}

\end{document}

BTW: It is much better to use, say \bigl( ... \bigr) instead of \big( ... \big), even if you do not notice the difference. Instead of & = 0.\int_{0}^{T} there should be & = 0\cdot\int_{0}^{T}.

Some comments and suggestions:

• Instead of a single align* environment, I'd use a nested equation/aligned combination. That way, you don't have to remember to the desired single equation number.

• Place & alignment markers at the beginning of each line, and shift lines 2 through 4 to the right via \quad (or \qquad) directives.

• The code you currently have for row 2, which features a single pair of \Bigg-sized absolute value bars, doesn't correspond to desired output suggested in the screenshot.

• I would exchange the order of terms in rows 3 and 4 to make them line up more easily with the material in rows 1 and 2.

• I recommend you load the mathtools package (a superset of the amsmath package) and use its \DeclarePairedDelimiter macro to create an \abs directive. That way, you can concentrate your coding on the *meaning * of the terms instead of on low-level symbol sizing.

A general comment: Your code doens't indicate that n (the number of terms in the sums) should be (must be?) a function of \Delta t (or, if you prefer, the number of partitions). Without such an indication, the result is trivial as there's a fixed, finite number of terms in the summation even as \Delta t\to0. Does the text that accompanies the equations explain that n is a suitably chosen function of \Delta t?

\documentclass{article}
\usepackage{geometry}  % set page parameters suitably
\usepackage{mathtools} % for '\DeclarePairedDelimiter' macro
\DeclarePairedDelimiter{\abs}{\lvert}{\rvert}

\begin{document}

\begin{equation} \label{fourth term 0}
\begin{aligned}[b]
&\lim_{\Delta t \to 0} \, \abs[\bigg]{\sum_{j=0}^{n-1}f_{tx}
   \bigl(t_j,X(t_j)\bigr)(t_{j+1} - t_j)\bigl(X(t_{j+1})-X(t_j)\bigr) } \\
&\quad\leq \lim_{\Delta t \to 0} \sum_{j=0}^{n-1} 
   \abs[\big]{f_{tx}(t_j,X(t_j))} (t_{j+1} - t_j) \cdot
   \abs[\big]{X(t_{j+1})-X(t_j)} \\
&\quad\leq
   \biggl[ \lim_{\Delta t \to 0}\sum_{j=0}^{n-1} \, 
   \abs[\big]{f_{tx}(t_j,X(t_j)) } 
   (t_{j+1}-t_j) \biggr]
   \cdot \lim_{\Delta t \to 0\,} \max_{\,0 \leq k \leq n-1} 
   \abs[\big]{X(t_{k+1}-X(t_k)} \\
&\quad= \int_{0}^{T}\abs[\big]{ f_{tx}(t,X(t)) }\,dt \cdot 0 = 0\,.
\end{aligned} \end{equation}

\end{document}