Lebesgue integrable implies measurable. Suppose we define Lebesgue integral like this, but without assuming that $f$ is measurable. Suppose there exists a finite integral $\int_Xfd\mu$. Does it follow that $f$ is measurable?
UPD. $f$ is integrable, if $\Xi(f) = \xi(f)$.
 A: An attemp to develop Behnam Esmayli's idea.
$f$ is Lebesgue integrable $\implies (\forall\epsilon>0)(\exists P_{\epsilon}: E = \sqcup_{i=1}^{\infty}E_i)\quad U(P_{\epsilon}, f) - L(P_{\epsilon}, f) < \epsilon$, where
$U(P_{\epsilon},f) = \sum_{i=1}^{\infty}\sup_{E_i}f(x)\mu(E_i), \quad L(P_{\epsilon},f) = \sum_{i=1}^{\infty}\inf_{E_i}f(x)\mu(E_i)$.
Define for $\epsilon = 1, \frac{1}{2}, ..., \frac{1}{n}, ... \quad h_n(x) = \sum_{i=1}^{\infty}\inf_{E_i}f(x)*\chi_{E_i}(x)$, $h_n(x) \leq h_{n+1}(x) \leq f(x)$, $h_n(x)$ is measurable.
Similarly $g_n(x) = \sum_{i=1}^{\infty}\sup_{E_i}f(x)*\chi_{E_i}(x).$ Then $(\exists N \in \mathbb {N})(\forall x \in E) \quad f(x) -h(x) < g(x) - h(x) < \frac{1}{n}$, thus $h_n(x) \rightrightarrows f(x)$, and since $h_n(x)$ is measurable, $f(x)$ is also measurable.
Any critics, please.
A: So if you can construct a non-measurable set $A \subset (0,1)$ and a function
$f(x)= 1$ for $x \in A$
$f(x)=-1$ for $-x \in A$
$f(x)=0$ $otherwise$
over the interval [-1,1]. f is an odd function so if you can define the integral without measurable f this has a good chance of having a defined integral of 0 over [-1,1].
If your definitions force the preimage of 1 to be measurable it is likely those same restrictions will force f to be measurable.
