Let $(X, \mathcal{A} ,\mu)$ measure space and $f:X \mapsto [0,+\infty]$ an integrable function. Define

$f_n(x) = \begin{cases} f(x) & \text{if} \ \ \ f(x) <n \\ n & \, \text{otherwise} \end{cases}$

Show that $f_n \rightarrow f$ in $L^1$.

My solution so far $\int |f_n-f|d\mu=\int \limits_{[f<n]}|f_n-f|d\mu+\int \limits_{[f\geq n]}|f_n-f|d\mu=0 \cdot\mu([f_n<n])+(f(x)-n) \cdot \mu([f_n \geq n])\rightarrow 0$

since $\mu([f_n \geq n]) \rightarrow 0$, because $f$ is integrable. I think I am missing something.


1 Answer 1


You may just use dominated convergence. First notice that $f_n(x) \leq f(x)$ for all $x \in X$ by definition. Further you have $f_n \to f$ pointwise for $n \to \infty$. Finally by assertion you have $f \in L^1(X)$ and thus dominated convergence yields that $\Vert f_n - f \Vert_1 \to 0$ for $n \to \infty$.

Nonetheless you attempt is wrong, since $$\int \limits_{[f\geq n]}|f_n-f|\, d\mu= \int \limits_{[f\geq n]} (f(x)-n) \, d\mu(x) = \int \limits_{[f\geq n]} f(x) \, d\mu(x) - n \cdot \mu([f \geq n]).$$ Your equality doesn't make sense like it is written there above.

  • $\begingroup$ Just saw that! But is it true that $f_n(x) \leq f(x)$ for all $x \in X$ or is it $f_n(x) \leq n$? (Still $f_n \rightarrow f$ and we use dominated convergence - thanks for the tip) $\endgroup$
    – dr_kelly
    Commented Feb 4, 2018 at 15:44
  • $\begingroup$ Just look at the two cases: 1. Case: $x \in X$ such that $f(x) < n$, in this case we have $f_n(x) = f(x)$. 2. Case: $x \in X$ such that $f(x) > n$, then we have $f_n(x) = n < f(x)$. Thus it follows $f_n(x) \leq f(x)$ for all $x \in X$. Please check my answer if you have no futher doubt :) $\endgroup$
    – Yaddle
    Commented Feb 4, 2018 at 18:00

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