integral over a subset of interval in $\mathbb{R}$ Consider a finite interval  $[0,d]$, where $d$ is a positive real number.
Let $K$ be a measurable subset of $[0,d]$
Then, how can I prove or disprove that $\int_Kx \,dx \geq \int^{m(K)}_0 x\,dx$, where $m(\cdot)$ denotes the Lebesgue measure?
 A: \begin{equation}
\begin{split}
\int_K x \,dx &=\int_{K\cap [0,m(K)]} x\,dx + \int_{K\cap [m(K), d]} x \,dx\\
&\ge \int_{K\cap [0,m(K)]} x\,dx + m(K) m(K \cap [m(K), d])  \ \ \ \ \ \ \ (1)\\
&= \int_{K\cap [0,m(K)]} x\,dx + m(K) m([0, m(K)] \setminus K)\ \ \ \ \ \ \ (2) \\
&\ge \int_{K\cap [0,m(K)]} x\,dx + \int_{[0, m(K)]\setminus K} x \,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)\\
&= \int_0^{m(K)} x\,dx.
\end{split}
\end{equation}
Remark
(1): As $x\ge m(K)$ on $K\cap [m(K), d]$, 
$$ \int_{K\cap [m(K), d]} x \,dx \ge  \int_{K\cap [m(K), d]} m(K) \,dx = m(K) m(K\cap [m(K), d])$$
(2) Note that $K \subset [0,d] = [0,m(K)] \cup [m(K), d]$, so 
$$m(K) = m(K \cap [0, m(K)]) + m(K \cap [m(K), d])$$
On the other hand, 
$$m(K) = m([0,m(K)]) = m(K\cap [0, m(K)]) + m([0,m(K)]\setminus K)$$
This two equalities implies 
$$m(K \cap [m(K), d]) = m([0,m(K)]\setminus K).$$
(3) Similar as (1), as $x \le m(K)$ on $[0,m(K)]$, we have 
$$\int_{[0, m(K)]\setminus K} x \,dx \le \int_{[0, m(K)]\setminus K}  m(K) \,dx = m(K) m([0,m(K)]\setminus K)$$
