Let $f$ a non-negative Lebesgue integrable function $ ( f\in L^+([0 ,\infty)) )$ such that $\displaystyle{ \lim_{x \to +\infty} f(x)}$ exists (finite or infinity). Prove that $\displaystyle{\lim_{x \to +\infty} f(x) =0}$.
Here it is the only thing I did.
Consider an increasing sequence $ (f)_n$ of simple non-negative functions such that $ f_n \to f$ pointwise. Then $ \int f =\lim \int f_n$.
1st case: $\displaystyle{\lim_{x \to +\infty} f(x) = l < +\infty}$.
Let $ \epsilon >0$, then there exists $ r>0 $ such that $ |f(x)-l|< \epsilon , \quad \forall x>r$.
Best Answer
Argue by contradiction: There are only two other possible cases. Suppose $ \displaystyle \lim_{x\to \infty} f(x) = L $ where $ L \in (0,\infty) .$ Then by definition of a limit, there exists $ a \in \mathbb{R}^+ $ such that $ f(x) > L/2 $ for all $ x> a.$ Estimating the integral: $$ \int^{\infty}_0 f(x) dx = \int^a_0 f(x) + \int^{\infty}_a f(x) dx > \int^a_0 f(x) dx + \frac{L}{2} \int^{\infty}_a 1 dx = \infty$$
which contradicts the assumption that $f\in L^1.$ Now you try to construct a similar argument for the case $L=\infty.$