Show $\mathbb{E}[X] \le \mathbb{E}[Y]$ I have a problem that I'm having trouble to identify the correct integral intervals for proof. I would like to have some help on the thought process.

Problem:
  Let $X$ and $Y$ have densities $f$ and $g$, respectively, and 
  $f(x)  
\begin{cases}
\ge g(x),  & \text{if $x$ }\le a;\\
\le g(x), & \text{if $x$ }\ge a
\end{cases}
$
Show that $\mathbb{E}[X] \le \mathbb{E}[Y]$

Since $\mathbb{E}[X]= \displaystyle  \int_{-\infty}^{\infty} xf(x)\, dx$
I will have to identify intervals (one positive and one negative) that are able to prove $\mathbb{E}[X] \le \mathbb{E}[Y]$ for all cases. However, I feel there may be an intersection, which leads to 3 intervals total, but I'm not sure how to divide up the segments. I would like some help with laying out the proof.
 A: W.l.o.g. let be $a=0$ (otherwise shift the random variables, see Addendum). Then 
\begin{align}
\int_{-\infty}^\infty xf(x)\,dx & = \int_{-\infty}^0 xf(x)\,dx + \int_0^\infty xf(x)\,dx \\[12pt]
& \leq \int_{-\infty}^0 xg(x)\,dx + \int_0^\infty xg(x)\,dx = \int_{-\infty}^\infty xg(x)\,dx
\end{align}
holds.
Addendum (maybe some extra notes about the "$a=0$" assumption above do not harm): With the linearity of the expectation value the following equivalence holds
$$\Bbb E(X-a)\leq \Bbb E(Y-a) \Leftrightarrow \Bbb E(X)-a\leq \Bbb E(Y)-a \Leftrightarrow \Bbb E(X)\leq \Bbb E(Y),$$
which justifies just to consider the translated variables.
For the CDF of $X-a$ (and for $Y-a$ analogously) we have
$$P_X(X-a\leq x) = P_X(X\leq x+a) = \int_{-\infty}^{x+a}f(x)\,dx = \int_{-\infty}^x f(x+a)\,dx.$$
For the translated desnsity function $f(x+a)$ (this is the density function of $X-a$) we obtain now
$$f(x+a)
\begin{cases}
\ge g(x+a),  & \text{if $x$ }\le 0\\
\le g(x+a), & \text{if $x$ }\ge 0
\end{cases}.$$
This allows us just to study the case $a=0$.
