# Lebesgue integral and limit

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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On a related note, it's a good exercise to construct an integrable function for which $\lim_{x \to +\infty} f(x)$ does not exist. –  Nate Eldredge Feb 2 '12 at 14:08
@NateEldredge: $\displaystyle{ f:= \chi_{\mathbb{Q}}}$ is an integrable function , $\displaystyle{ \int f=0}$ and $\displaystyle{\lim_{x \to \infty} f(x)}$ does not exist. –  passenger Feb 2 '12 at 14:11
Yep, that works. Now can you find one which is continuous? –  Nate Eldredge Feb 2 '12 at 14:33
@NateEldredge: This is more difficult... I have to think about it –  passenger Feb 2 '12 at 14:42
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...?) –  AD. Feb 2 '12 at 16:10

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 $L=\infty.$
Thank's for the answer! For the case $L=\infty$ I said that for all $\epsilon>0$ $\exists r >0 : f(x) > 1/\epsilon , \forall x\geq r$ and continue the same as you. I have one question. In your solution where do you use properties of Lebesgue integral (as for example did in beginning ) I mean this solution uses only properties of Riemann integral. –  passenger Feb 2 '12 at 14:20
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. –  Ragib Zaman Feb 2 '12 at 14:33