Let $\phi$ be a convex function on $(-\infty, \infty)$, $f$ a Lebesgue integrable function over $[0,1]$ and $\phi\circ f$ also integrable over $[0,1]$. Then we have:
$$\phi\Big(\int_{0}^{1} f(x)dx\Big)\leq\int_{0}^{1}\Big(\phi\circ f(x)\Big)dx.$$
I am thinking about a counterexample that the converse of this statement. In other words, I am trying to find a $\phi$ which is convex on $\mathbb{R}$, and $f$ is a Lebesgue integral function (on some set), satisfying $\phi\Big(\int_{0}^{1} f(x)dx\Big)\leq\int_{0}^{1}\Big(\phi\circ f(x)\Big)dx$, but $\phi\circ f$ is not Lebesgue integrable (on some set).
I tried to use the convexity of non-integrability of $\dfrac{1}{x}$. However, $\dfrac{1}{x}$ is not convex in the whole $\mathbb{R}$, so I tried to use $\left|\dfrac{1}{x}\right|$. So define $\phi:=\left|\dfrac{1}{x}\right|$.
We know that $f(x)=x$ is Lebesgue integrable, and if we restrict our case to $\mathbb{R^{+}}$, then $\phi\circ f=\dfrac{1}{x}$ which is not Lebesgue integrable.
Then, we have $$\phi\Big(\int_{1}^{2} f(x)dx\Big)=\dfrac{2}{3}<\int_{1}^{2}\Big(\phi\circ f(x)\Big)dx=\log(2).$$
Is my argument correct? I feel that I am kind of in a self-contradiction, or my attempt to show the converse of the statement of Jensen's inequality is wrong from the beginning.
Thank you so much for any ideas!
Let me restate my question here.
In fact, I need to find a $\phi(x)$ which is convex on whole $\mathbb{R}$, $f(x)\in L^{1}(\mathbb{R})$ and $\phi(\int f)\leq\int\phi(f)$, but $\phi(f)\notin L^{1}(\mathbb{R})$.
In my post before, I used $f(x)=x$, but I realized that $f(x)$ is no integrable over $\mathbb{R}$, so I modified it a little bit and now here is a valid counter-example.
Consider $\phi(x)=\Big|\dfrac{1}{x}\Big|$ and $f(x)=x$ if $x\in[0,1]$ but $f(x)=0$ at all other $x$. It is clear that $\phi(x)$ is convex on $\mathbb{R}$, and $f(x)\in L^{1}(\mathbb{R})$.
Now, $\phi(\int_{0}^{1}f(x)dx)=2$, but $\int_{0}^{1}\phi(f(x))dx=\infty$.
Thus, we have $\phi(\int_{0}^{1}f(x)dx)<\int_{0}^{1}\phi(f(x))dx$ but $\phi(f(x))$ is not Lebesgue integrable over $[0,1]$ by definition, and thus it cannot be integrable over $\mathbb{R}$.