What is the intuition behind Chebyshev's Inequality in Measure Theory

Solution 1:

It can be helpful to draw a picture:

Here:

• the blue curve is $f(x)$,
• the base of the red box is the set $\{x \in E: f(x) \ge \lambda\}$,
• the height of the red box is $\lambda$.

Chebyshev's inequality says that the area in the red box is less than the area under the blue curve $f(x)$.

The only issue with this picture is that, depending on $\lambda$ and $f$, you might have multiple boxes under the curve at different locations, instead of just one. But then the same thing applies to the sum of the areas under the boxes.

Solution 2:

The essential point is that $$0 \leq \lambda \cdot 1_{\{x \in E \;|\; f(x) \geq \lambda \} } \leq f$$

where $1_{\{x \in E \;|\; f(x) \geq \lambda \} }$ is the characteristic (indicator) function of $\{x \in E \;|\; f(x) \geq \lambda \}$.

Let us see.

Let $A$ be $\{x \in E \;|\; f(x) \geq \lambda \} $. Then it is clear that $$0 \leq \lambda \cdot 1_A \leq f$$ So $$0 \leq \lambda \cdot m(A)= \int_E \lambda \cdot 1_A \leq \int_E f$$

So, since $\lambda >0$, we have

$$m(\{x \in E \;|\; f(x) \geq \lambda \} )= m(A) \leq \frac{1}{\lambda}\int_E f $$

Solution 3:

You could think of it like this. At a birthday party, everyone eats a certain amount of cupcakes. The total number of cupcakes eaten is greater than or equal to $\lambda$ times the number of people who ate at least $\lambda$ cupcakes.

Translating this into a proof: Let $S = \{ x \in E \mid f(x) \geq \lambda \}$. Then $$ \int_E f \geq \int_S f \geq \int_S \lambda = \lambda m(S). $$

Solution 4:

Flip it around, and cutting E for simplicity, we get:

$$m\{ f(x) \ge \lambda \} * \lambda \leq \int f$$

The measure over the region where $f$ is at least $\lambda$, times $\lambda$, isn't bigger than the integral over the entire region.

Imaging we take a our function, and we define a cut off we call $\lambda$. We shave off everything higher than $\lambda$, and we set everything under $\lambda$ to zero. Chebychev simply states we didn't make its integral bigger by doing this.

Let $g_S$ be the function 1 on $S$ and 0 elsewhere. Then $\lambda g_S$ is 0 everywhere except on some set $S$, where it equals $\lambda$. The integral is equal to the measure of $S$ times $\lambda$.

This $g = \lambda g_{f(x)\ge\lambda}$ is (non-strictly) under $f$, and the set where it is non-zero is exactly your set $A$, where it has height $\lambda$. Its integral is $\lambda$ times the measure of the set it is non-zero. The difference between them, $h = f-g$ is non-negative, and $\int f = \int (g+h)$. $h$ is non-negative, so you'd expect adding it to $g$ would make its integral not-smaller.

$\int g$ is basically our $m\{f(x) \ge \lambda\} * \lambda$ here, and $h$ is the excess portion of $f$ above $g$.