Meaning of $f(x, \cdot)$ Let $(X,Σ, \mu)$ be a measure space and $(a,b)$ non empty open interval in $\mathbb{R}$. Let $f:X \times (a,b) \to \mathbb{C}$ be a function and suppose that the integral $F(t) = \int f(x,t) dx$ is defined for every $t \in (a,b)$.
The question is: 

Using Lebesgue dominated convergence theorem how can we make the function $F(t)$ continuous? 

This is the proof. 

Suppose for almost every $x \in X$ the function $f(x,\cdot)$ is continuous in $(a,b)$ and there is an integrable function $g:X \to [0,\infty)$ such that $|f(x,t)| \leq g(x)$ for every $t \in (a,b)$ and almost every $x \in X$. Then F is continuous in $(a,b)$.

I understand the proof but I do not understand the notation $f(x, \cdot)$. What does  $f(x, \cdot)$mean? 
 A: It is given that $$f(\cdot, \cdot): X \times (a,b)\to \mathbb C$$ That is $f$ has two arguments, $x$ that takes values in $X$ and $t$ that takes values in $(a,b)$. Hence, if you fix $x$ in $X$ then $f(x, \cdot): (a,b) \to \mathbb C$ is a function of $t$.
A: $f(x,\cdot)$ is the function $g:(a,b)\to\mathbb{C}$ given by $g(t)=f(x,t)$. Notice that $g$ depends on $x$. The notation $f(x,\cdot)$ makes this dependence explicit. This dependence is lost if you write $f(t)$; but $f_x(t)$ is ok. Indeed, W. Rudin uses $f_x$ instead of $f(x,\cdot)$ in his book Real and Complex Analysis.
The advantage of using $f(x,\cdot)$ is that we don't need make a new definition explicitly. If we want to use $f_x$ we have to define it explicitly (because it's not clear from the context what it is).
Example: if $f(x,y)=x+y$ then $f(5,\cdot)$ is the function $g(t)=5+t$.
A: In this context for every fixed $x\in X$ we have a function $(a,b)\to\mathbb C$ that is prescribed by: $$t\mapsto f(x,t)$$
$f(x,\cdot)$  is a notation for this function.
