Markov's Inequality

• categories: Measure Theory, Theorem

Let $(X,\mathcal{A},\mu )$ be a measure space, and let $f:X\to [0,\infty )$ be a non-negative measurable function. For any $\alpha >0$, the inequality states:

$$\mu (\{x\in X:f(x)\ge \alpha \})\le \frac{1}{\alpha }{\int }_{X}fd\mu .$$

Intuition:

Markov’s inequality tells us that the “size” (measure) of the set where a function $f$ exceeds a certain threshold $\alpha$ is bounded by the average value (integral) of $f$ divided by $\alpha$.

Proof:

1. Define the set: Let $A_{\alpha }=\{x\in X:f(x)\ge \alpha \}$ be the set where $f(x)$ exceeds the threshold $\alpha$. We want to find an upper bound for $\mu (A_{\alpha })$.

2. Bound $f$ on the set $A_{\alpha }$: By definition of $A_{\alpha }$, for all $x\in A_{\alpha }$, we have $f(x)\ge \alpha$. Thus, we can write:

   $$f(x)\ge \alpha \cdot {\mathbb{I}}_{A_{\alpha }}(x),$$

   where ${\mathbb{I}}_{A_{\alpha }}(x)$ is the indicator function of the set $A_{\alpha }$, which is $1$ if $x\in A_{\alpha }$ and $0$ otherwise.

3. Integrate both sides: Now integrate both sides over the entire space $X$ with respect to the measure $\mu$:

   $${\int }_{X}f(x)d\mu (x)\ge {\int }_X\alpha \cdot {\mathbb{I}}_{A_{\alpha }}(x)d\mu (x).$$

4. Simplify the right-hand side: Since ${\mathbb{I}}_{A_{\alpha }}(x)$ is $1$ only on the set $A_{\alpha }$ and $0$ elsewhere, the integral on the right-hand side simplifies to:

   $${\int }_X\alpha \cdot {\mathbb{I}}_{A_{\alpha }}(x)d\mu (x)=\alpha \cdot \mu (A_{\alpha }).$$

5. Final inequality: Therefore, we have:

   $${\int }_{X}f(x)d\mu (x)\ge \alpha \cdot \mu (A_{\alpha }).$$

6. Rearrange to obtain Markov’s inequality: Dividing both sides by $\alpha$ (where $\alpha >0$), we obtain:

   $$\mu (A_{\alpha })=\mu (\{x\in X:f(x)\ge \alpha \})\le \frac{1}{\alpha }{\int }_{X}f(x)d\mu (x).$$