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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$.

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    Hint to Nates exercise: 1) Start to think about a convergent series $A= \sum a_n$. 2) One way to construct a function $f$ such that $\int f = A$ is to put $f(x)=a_n$ on the interval $I_n=(n,n+1)$. Can you find an other way? (Note that $|I_n|=1$, fairly thick...?)2012-02-02

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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 $L=\infty.$

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    We need the condition that $f \in L^1$ for the integrals in the proof to even make sense, and then we invoke it at the end for the contradiction, but not so much other than that.2012-02-02