Cumulative Distribution Functions and Continuity The CDF is defined as $$F(x) = \int_{-\infty}^{x} f$$
Where $f$ is the pdf for a random variable $X$. I have been taught that if $X$ is a continuous random variable, then $P(X = b) = 0$. 
I present the following example



According to wikiapedia, if I want to find $P(X = 1)$, I use
$$P(X = b) = F(b) - \lim_{x \to b^-} F(x)$$
According to the formula above, it is equivalent to
$$P(X = b) = \int_{-\infty}^{b} f - \lim_{x \to b^-} \int_{-\infty}^{x} f$$
With reference to the pictures (or not?), does the above equation actually mean we are finding the area of a very small rectangle? 
That is $\int_{-\infty}^{1} f$ gives us the "complete" area of the triangle at $x = 1$ and $\lim_{x \to 1^-} \int_{-\infty}^{x} f$ gives us the "same" area, but just a bit to the left of $x = 1$. Their difference gives us a very thin rectangle
 A: Note that the Wikipedia formula:

$$
P(X = b) = F(b) - \lim_{x \to b^-} F(x)
$$

only really comes in handy if the distribution of $X$ is discontinuous at $b$. You'll notice, however, that your edited CDF is now continuous everywhere, which means that for any $b\in \Bbb{R}$:
$$
P(X = b) = F(b) - \lim_{x \to b^-} F(x) = F(b) - F(b) = 0
$$
which makes sense, since $X$ is a continuous random variable.
A: The PDF of a continuous random variable need not be continuous, only its CDF.
Latest news! The OP is in fact interested in the CDF $F$ such that $F(x)=\frac12x$ if $0\leqslant x\lt\frac14$, $F(x)=\frac32x$ if $\frac14\leqslant x\lt\frac12$, and $F(x)=1$ if $x\geqslant\frac12$. Thus $F$ has two jumps, each of height $\frac14$, at $x=\frac14$ and at $x=\frac12$. The sum of the sizes of these jumps is $\frac12\lt1$ hence the distribution of $X$ has also an absolutely continuous part, spread on $[0,\frac12]$, whose total mass is $1-\frac12=\frac12$. Thus, the distribution of $X$ is neither purely discrete nor purely absolutely continuous.
This phenomenon is hardly exotic (despite the efforts many presentations of the probabilistic apparatus make to hide it). Consider for example, what might be the simplest continuous random variable, namely, $U$ uniform on $[0,1]$. Then the distribution of $Y=\max(\frac13,U)$ is neither purely discrete nor purely absolutely continuous.
Likewise, to get some random variable $X$ with CDF $F$, consider $X=g(U)$, where $g(u)=2u$ if $0\leqslant u\lt\frac18$, $g(u)=\frac14$ if $\frac18\leqslant u\lt\frac38$, $g(u)=\frac23u$ if $\frac38\leqslant u\lt\frac34$, and $g(u)=\frac12$ if $\frac34\leqslant u\leqslant1$.
