integral over a subset of interval in $\mathbb{R}$

Question

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?

Answer 1

\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 $$\int_{[0, m(K)]\setminus K} x \,dx \le \int_{[0, m(K)]\setminus K} m(K) \,dx = m(K) m([0,m(K)]\setminus K)$$