Problem: Let $f$ be Riemann integrable on $[0,x]$ for all $x>0$. Prove that $$\liminf\limits_{x\to\infty}f(x)\leq\liminf\limits_{x\to\infty}\frac{1}{x}\int_{0}^{x}f(t)\,dt.$$
Thoughts: From the definition of $\liminf$, I need to prove that $$\lim\limits_{y\to\infty}\inf\limits_{x\geq y}f(x)\leq\lim\limits_{y\to\infty}\inf\limits_{x\geq y}\frac{1}{x}\int_{0}^{x}f.$$ Fix $y>0$. Then for any fixed $z\geq y$ it follows that $$\inf\limits_{x\in[y,z]}f(x)\leq\frac{1}{z}\int_{0}^{z}f.$$ Since this is true for all $z\geq y$, we have that $$\inf\limits_{z\geq y}\inf\limits_{x\in[y,z]}f(x)\leq \inf\limits_{z\geq y}\frac{1}{z}\int_{0}^{z}f.$$
Since the above holds for all $x>0$, it follows that $$\lim\limits_{y\to\infty}\inf\limits_{x\geq y}f(x)\leq\lim\limits_{y\to\infty}\inf\limits_{x\geq y}\frac{1}{x}\int_{0}^{x}f.$$
I am almost certain that I made a mistake handling all the infima, and that I failed to prove the result.
Could anybody please point out if I made a mistake. And if I did, could anyone give me a hint on how to look at the problem. Please don't solve it for me.
For $x>T>0$ we have $\frac 1 x \int_0^{x}f(t)dt =\frac 1 x \int_0^{T}f(t)dt+\frac 1 x \int_T^{x}f(t)dt \geq \frac 1 x \int_0^{T}f(t)dt+ g(T) \frac {x-T} x$ where $g(T)$ is the infimum of $f(t)$ for $t \geq T$. Now just take limit on both sides as $x \to \infty$. The first term tends to $0$ and the second term tends to $g(T)$ (which tends to $\lim \inf_{x \to \infty} f(x)$ as $T \to \infty$).