Let $(X,\mathcal{S},\mu)$ be a measure space with $\mu(X)=1$, and let $f\in \mathcal{L}^{1}(\mu)$. Show that $\mu(\{x: |f(x)-\int f \text{d}\mu|>c\})\leq\frac{1}{c^{2}}(\displaystyle\int f^{2}\text{d}\mu$ $(\int f\text{d}\mu)^{2})$ for $c>0$. This is known as Chebyshev's Inequality in probability theory. Hint: Apply Markov's inequality to $h(x)=(f(x)-\int f\text{d}\mu)^{2}$ and use that $\int 1 \text{d}\mu =1$.
My work so far:
Using the hint above, I get the following:
$\mu(\{x\in X: |(f(x)-\int f \text{d}\mu)^{2}|\geq c\})\leq \frac{1}{c}(\displaystyle\int |(f(x)-$ $\int f \text{d}\mu)^2|\text{d}\mu)= \displaystyle \int |f(x)^2-2f(x)$ $\int f(x)\text{d}\mu+ (\int f\text{d}\mu)^{2}|\text{d}\mu$. I'm not sure what to do from here or what the measure of $X$ being 1 has to do with anything. Any help would be welcome. Thank you.