Prove that $m(\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)\geq\frac{1}{2}\})\geq\frac{1}{2}$ under these conditions... Question: Suppose for each $j\in\mathbb{N}, f_j:[0,1]\rightarrow\mathbb{R}$ is Lebesgue measurable such that $0\leq f_j\leq\frac{3}{2}$ and $\int_0^1 f_j dm=1$.  Prove that $m(\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)\geq\frac{1}{2}\})\geq\frac{1}{2}$.
Thoughts/Attempt: Let $A=\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)\geq\frac{1}{2}\}$, and $B=\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)<\frac{1}{2}\}$.  Suppose, by contradiction, that $m(A)<\frac{1}{2}$.  So, we can split up the integral as $$\int_0^1f_jdm=\int_Af_jdm+\int_Bf_jdm$$ where we get equality by the integral in the assumption.  Now, $\int_Af_jdm<\frac{1}{2}$, by our (contradiction) assumption.  And, $\int_Bf_jdm<\frac{1}{2}$, using our set $B$.  Therefore, $$\int_0^1f_jdm<\frac{1}{2}+\frac{1}{2}=1$$ a contradiction, since this integral must equal $1$ from our assumption.  Hence, we contradict that $m(A)<\frac{1}{2}$.
However, I am not quite sure if this works, because our sets are dealing with the $\lim\sup f_j(x)$ as $x\in[0,1]$, but wouldn't I have to compensate in the integral since the image of $f_j$ is all of $\mathbb{R}$?
 A: For each $n\in\mathbb{N}$
$$\begin{align}
 1=\int f_n &=\int_{\{f_n\geq\frac12\}}f_n+\int_{\{f_n<\frac12\}}f_n\leq \frac32\lambda\Big(f_n\geq\frac12\Big) + \frac12\lambda\Big(f_n<\frac12\Big)\\
&=\lambda\Big(f_n\geq\frac12\Big)+\frac12
\end{align}$$
Hence
$$\lambda\Big(f_n\geq\frac12\Big)\geq\frac12\qquad\text{for all}\quad n\in\mathbb{N}$$
Notice that
$$\begin{align}
\Big\{\limsup_jf_j\geq\frac12\Big\}\supset\limsup_j\Big\{f_j\geq\frac12\Big\}=\bigcap_n\bigcup_{m\geq n}\Big\{f_m\geq\frac12\Big\}\tag{1}\label{one}
\end{align}$$
Putting things together we obtain
$$\lambda\Big(\limsup_jf_j\geq\frac12\Big)\geq\lim_n\lambda\Big(\bigcup_{m\geq n}\{f_m\geq\frac12\}\Big)\geq\limsup_n\lambda\Big(f_n\geq\frac12\Big)\geq\frac12$$

Comment:
The set inequality $\eqref{one}$ is clear since $x\in\bigcap_n\bigcup_{m\geq n}\Big\{f_n\geq\frac12\Big\}$ implies that $f_n(x)\geq\frac12$ infinitely many times, and so $\limsup_nf_n(x)\geq\frac12$.
A: To handle the $\limsup$ for your integrals over $B$, I suggest using Fatou's lemma.
Fatou's lemma only applies to non-negative functions, and $\tfrac 3 2 - f_n(x)$ is a non-negative function. Applying Fatou to $\tfrac 3 2 - f_n(x)$ on $B$, we have
$$ \int_B \liminf_{n \to \infty}\left(\tfrac 3 2 - f_n (x)\right) dm \leq \liminf_{n \to \infty} \int_B \left(\tfrac 3 2 - f_n (x)\right) dm,$$
or equivalently,
$$ \limsup_{n \to \infty} \int_B f_n(x) dm \leq \int_B \limsup_{n \to \infty} f_n(x) \leq \tfrac 1 2 m(B).$$
Meanwhile, the correct inequality for the integrals over $A$ is
$$ \int_A f_n(x) dm \leq \tfrac 3 2 m(A)$$
for each $n \in \mathbb N$. (It seems like you've missed the factor of $\tfrac 3 2$, coming from the upper bound on $f_n$.)
Thus
$$ 1 = \limsup_{n \to \infty}\int_0^1 f_n(x) dx \leq \tfrac 3 2 m(A) + \tfrac 1 2 m(B) =  m(A) + \tfrac 1 2,$$
which clearly implies that $m(A) \geq \tfrac 1 2$.
