Suppose a sequence of function $f_n \mapsto f$ converges in measure and in integral. by integral I mean $\int f_n \,d\mu \mapsto \int f \,d\mu$. Does that mean they converge in L1. It seems not entire true to me e.g. consider $f$ with compact support whose positive value cancels with the negatives and moves right $+ 1$ for each $n.$ Any hints?
2026-04-06 14:08:09.1775484489
Dose convergence in integral and measure imply convergence in L
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Consider $$f_n(x)=\begin{cases} n &\text{ if }x\in \big[0,\frac{1}{n}\big],\\ -n &\text{ if }x\in \big[-\frac{1}{n},0\big]\end{cases}$$ Then $f_n\xrightarrow{\text{ almost everywhere }\implies \text{ in measure}} 0$. Also, $0=\int_\Bbb R f_n\longrightarrow \int_\Bbb R0=0$, but $\int_\Bbb R|f_n-0|=2$.