Suppose that$ f \in L_1 (X)$, that $f (x) \gt 0$ for all $x \in X$, and that $0 \lt α \lt μ(X) \lt \infty$ Suppose that$ f \in L_1 (X)$, that $f (x) \gt 0$ for all $x \in X$, and that
$0 \lt α \lt μ(X) \lt \infty$. Prove that
inf
$\{\int_{E} f dμ : μ(E) \ge α\} \gt 0$
Give an example to show that the result fails if one drops the hypothesis $μ(X) \lt \infty$.
it is very hard problem for me. it is valuable for me to be solved.
 A: First, we will show the following:

Let $B = \{f \le t\}$ for some $t > 0$ and $A$ be an arbitrary set with $\mu(A) \ge \mu(B)$. Then $\int_A f \, d\mu \ge \int_B f \, d\mu$ holds.

To see this, note that $\mu(A^c \cap B) + \mu(A \cap B) = \mu(B) \le \mu(A)$, which implies $\mu(A^c \cap B) \le \mu(A) - \mu(A \cap B) = \mu(A \cap B^c)$. This means:
$$\int_{A^c \cap B} f \, d\mu \le \int_{A^c \cap B} t \, d\mu = t \mu(A^c \cap B) \le t \mu(A \cap B^c) = \int_{A \cap B^c} t \, d\mu \le  \int_{A \cap B^c} f \, d\mu$$
The assertion now follows:
$$\int_B f \, d\mu = \int_A f \, d\mu + \int_{A^c \cap B} f \, d\mu - \int_{A \cap B^c} f \, d\mu \le \int_A f \, d\mu$$
To prove the original problem, note that the sequence $M_n = \{f \le \frac{1}{n}\}$ is a decreasing sequence with $\bigcap_n M_n = \emptyset$. Since $\alpha > 0$, we can find a positive integer $n$ that satisfies $\mu(M_n) < \alpha$ (finite measures are continuous from above).
If $\mu(M_n) = 0$, then $f$ is almost surely bounded from below by $\frac{1}{n}$. But this means $\int_E f \, d\mu \ge \frac{\mu(E)}{n} \ge \frac{\alpha}{n}$ for any set $E$ with $\mu(E) \ge \alpha$.
If $\mu(M_n) > 0$, we can apply the lemma. Then for any set $E$ with $\mu(E) \ge \alpha > \mu(M_n)$ we have $\int_E f \, d\mu \ge \int_{M_n} f \, d\mu > 0$.

For the counterexample, consider e.g. the Lebesgue-measure with $f(x) = \exp(-|x|)$.
A: Proceed by contradiction and assume that $\int_{E_n} f d\mu \to 0.$ Since $f$ is integrable and $X$ has finite measure, convergence in measure shows at once $\mu(|f \mathbf{1}_{E_n}| \leq \varepsilon) \to \mu(X).$ So, for any $\varepsilon > 0$ there exists some $n_\varepsilon \in \mathbf{N}$ such that $\mu(|f \mathbf{1}_{E_n}| \leq \varepsilon) > \alpha$ for all $n \geq n_\varepsilon.$ Next,
$$
\mu(|f \mathbf{1}_{E_n}| \leq \varepsilon) = \mu(E_n \cap \{f \leq \varepsilon\}) \leq \mu(f \leq \varepsilon).
$$
We now show that $\lim\limits_{\varepsilon \downarrow 0} \mu(f \leq \varepsilon) = 0.$ Indeed,
$$
\mu(f \leq \varepsilon) = \mu(X) - \mu(f > \varepsilon)
$$
and we may apply dominated convergence theorem (since $X$ has finite measure) to obtain
$$
\mu(f > \varepsilon) \to \mu(f > 0) = \mu(X).
$$
Finally, for $\varepsilon > 0$ small enough, $\mu(f \leq \varepsilon) < \dfrac{\mu(X)}{2}$ and thus, this contradicts $\mu(|f \mathbf{1}_{E_n}| \leq \varepsilon) \to \mu(X).$ QED
For the example, consider any disjoint sets $E_n,$ define $X$ to be their union and assign mass 1 to each of the $E_n$ and $f$ to be $\frac{1}{n^2}$ on $E_n.$ Then, $1 < \mu(E_n)$ for every $n,$ $f \in \mathscr{L}_1$ and yet $\inf_n \int_{E_n} f = 0.$
A: Lemma 1: Let $(A_{n})$ be a sequence of subsets of $X$. Then $1_{\limsup A_{n}}=\limsup_{n}1_{A_{n}},$
where $\limsup A_{n}=\cap_{n\in\mathbb{N}}\cup_{k\geq n}A_{k}$.
Proof of Lemma 1: Let $x\in X$. Case 1: $x\in\limsup A_{n}$, then
$x\in A_{n}$ for infinitely many $n$ Choose $n_{1}<n_{2}<n_{2}<\ldots$
such that $x\in A_{n_{k}}$ for $k=1,2,\ldots$. It follows that $1_{A_{n_{k}}}(x)=1$
for all $k$, so $\limsup_{n}1_{A_{n}}(x)\geq1$. On the other hand,
$1_{A_{n}}(x)\leq1$ for all $n$, so $\limsup_{n}1_{A_{n}}(x)\leq1$.
Therefore, $\limsup_{n}1_{A_{n}}(x)=1=1_{\limsup A_{n}}(x)$.
Case 2: $x\notin\limsup A_{n}$, then there exists $N$ such that
$x\notin A_{k}$ for all $k\geq N$. Therefore, for any $n\geq N$,
$\sup_{k\geq n}1_{A_{k}}(x)=0$ and hence $\limsup_{n}1_{A_{n}}(x)=\lim_{n\rightarrow\infty}\sup_{k\geq n}1_{A_{k}}(x)=0$.
This shows that $1_{\limsup A_{n}}(x)=\limsup1_{A_{n}}(x)=0$.

Now, we go back to your question. Prove by contradition. Suppose the
contrary that the infimum is zero, then there exist a sequence of
measurable sets $(E_{n})$ such that $\mu(E_{n})\geq\alpha$ and $\int_{E_{n}}fd\mu<\frac{1}{2^{n}}$.
Let $E=\limsup_{n}E_{n}$. By Dominated Convergence Theorem,
\begin{eqnarray*}
\int_{E}fd\mu & = & \int1_{\limsup E_{n}}fd\mu\\
 & = & \int\limsup1_{E_{n}}fd\mu\\
 & = & \lim_{n\rightarrow\infty}\int\sup_{k\geq n}1_{E_{k}}fd\mu\\
 & \leq & \lim_{n\rightarrow\infty}\int\sum_{k=n}^{\infty}1_{E_{k}}fd\mu\\
 & = & \lim_{n\rightarrow\infty}\sum_{k=n}^{\infty}\int_{E_{k}}fd\mu\\
 & \leq & \lim_{n\rightarrow\infty}\sum_{k=n}^{\infty}\frac{1}{2^{k}}\\
 & = & 0.
\end{eqnarray*}
Since $f>0$ everywhere, it follows that $\mu(E)=0$ (This is a well-known
fact, so I do not prove it. Consider $A_{n}=\{x\in E\mid f(x)>\frac{1}{n}\}$...).
On the other hand, since $\mu(X)<\infty$, by continuity of measure,
\begin{eqnarray*}
0 & = & \mu(E)\\
 & = & \lim_{n\rightarrow\infty}\mu(\cup_{k\geq n}E_{k})\\
 & \geq & \lim_{n\rightarrow\infty}\mu(E_{n})\\
 & \geq & \alpha
\end{eqnarray*}
which is a contradiction.
Remark: The condition that $\mu(X)<\infty$ cannot be dropped. Otherwise,
consider $([1,\infty),\mathcal{B}([1,\infty),m)$ and $f(x)=\frac{1}{x^{2}}$...
