Prove that if a function $f$: $\mathbb{R}$ $\longrightarrow$ $\mathbb{R}$ is finite almost everywhere then its lebesgue integral is not nessecerily finite:
I considered the characteristic function $X_{[0,\infty]}$ which is finite everywhere but its lebesgue integra is infinite..
But i want to find set $E\neq\emptyset$ in which f is infinite and $m(E)=0$.
So i give this counterexample which may be silly:
I took $f(x)=\infty$ if x is rational and $f(x)=1$ if x is irrational.
Is this example right?
If not ,can someone give me a correct example?
Considere $\begin {align} f:& \mathbb R \rightarrow \mathbb R\\&x \mapsto \frac 1 {x^2} \space\mathrm {if}\space x \ne 0\end {align}, f(0)=\infty$. It is trivially finite on $\mathbb R^*$ so almost everywhere.
But its Lebesgue integral on $[-1, 1]$, nor on any segment containing $0$ is not because its antiderivative is $- \frac 1 x$ which has an infinite limite in $0$.
The rational behind this is that being finite almost everywhere of does not imply being bounded. The measure of the set $\{ x \in \mathbb R / f(x) > A \}$ is $\frac 2 {\sqrt A}$, so its limit will be 0 for $A \rightarrow \infty$, but the surface below the function is greater than $\frac 2 {\sqrt A} A=2 \sqrt A$ and the limit will be $\infty$
Your example was not correct, because $f(0) = 0$ except on a set of measure 0, so the Lebesgue integral will be 0 on any interval...