Let $(X, \mathcal B, m)$ be a measure space. For $1 \leq p < q \leq \infty$, under what condition is it true that $L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m)$ and what is a counterexample in the case the condition is not satisfied?

  • $\begingroup$ Do you want $L^q(X,\mathcal B,m)$ to be any subset of $L^p(X,\mathcal B, m)$, or a proper subset? Also did you mean $p \leq q$ or something else in the question? $\endgroup$
    – Srivatsan
    Commented Sep 20, 2011 at 10:23
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    $\begingroup$ en.wikipedia.org/wiki/Lp_space#Embeddings $\endgroup$
    – user16391
    Commented Sep 20, 2011 at 10:23
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    $\begingroup$ You should also consider the counting measure on a finite set - what happens in this case? $\endgroup$ Commented Sep 20, 2011 at 12:25
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    $\begingroup$ The question has been effectively answered by the answers below in the case of finite spaces. For infinite spaces it may be interesting to read: math.stackexchange.com/questions/55170/… $\endgroup$ Commented Sep 20, 2011 at 12:50
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    $\begingroup$ Related: math.stackexchange.com/questions/1371017/… $\endgroup$
    – Watson
    Commented Jun 6, 2016 at 16:57

7 Answers 7


Theorem Let $X$ be a finite measure space. Then, for any $1\leq p< q\leq +\infty$ $$L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).$$ The proof follows from Hölder inequality. Note that $\frac{1}{p}=\frac{1}{q}+\frac{1}{r}$, with $r>0$. Hence $$\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.$$

The case reported on the Wikipedia link of commenter answer follows from this, since of course, if $X$ does not contain sets of arbitrary large measure, $X$ itself can't have an arbitrary large measure.

For the counterexample: $f(x)=\frac{1}{x}$ belongs to $L^2([1,+\infty))$, but clearly it does not belong to $L^1([1,+\infty)).$


I would like to add other lines to this interesting topic. Namely, I would like to prove what is mentioned in Wikipedia, hope it is correct:

Theorem Suppose $(X,\mathcal B,m)$ is a measure space such that, for any $1\leq p<q\leq +\infty,$ $$L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m).$$ Then $X$ doesn't contain sets of arbitrarily large measure.

Indeed it is well defined the embedding operator $G:L^q(X, \mathcal B, m) \to L^p(X, \mathcal B, m)$, and it is bounded.

Indeed the inclusion $L^q(X,\mathcal B,m)\subset L^p(X,\mathcal B,m)$ is continuous. Convergence in $L^p$ and in $L^q$ imply convergence almost everywhere and we can conclude by the closed graph theorem.

By Hölder inequality, $$\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}.$$ This means $$\|G\|\leq \text{meas}(X)^{1/r}=\text{meas}(X)^{1/p-1/q}.$$ But, considering $f(x)=\chi_X(x)$, one sees that $$\|G\|=\text{meas }(X)^{1/r}<+\infty \Leftrightarrow \text{meas }(X)<+\infty.$$ Now we can proceed by density of the vector space of the simple functions in both $L^p(X,\mathcal B,m)$ and $L^q(X,\mathcal B,m).$

Theorem Let $(X,\mathcal B,m)$ be a measure space. Then $X$ doesn't contain sets of arbitrarily small measure if and only if for any $1\leq p<q\leq +\infty$, one has $$L^p(X, \mathcal B, m) \subset L^q(X, \mathcal B, m).$$

Let us suppose that, for any subset $Y\subseteq X,\quad Y\in\mathcal B$, we have $0<\alpha\leq\text{meas}(Y)$.

It sufficies to prove the statement for simple functions. Pick now $$f(x) =\sum_{j=1}^n a_j\chi_{E_j},$$ where $\{E_j\}_{j=1,\dots,n}$ is a collection of disjoint subsets of $\mathcal B.$ Then $$\|f\|_{L^q} \le \sum_{j=1}^n \|a_j\text{meas}(E_j)\|_{L^q} = \sum_{j=1}^n a_j\text{meas}(E_j)^{1/q}=\sum_{j=1}^n a_j\text{meas}(E_j)^{1/q-1/p+1/p}\leq\frac{1}{\alpha^{1/p-1/q}}\|f\|_{L^p}.$$
The first inequality is due to Minkowski inequality.

For the converse of the theorem note that again it is well defined the embedding operator $G:L^p(X,\mathcal B,m)\to L^q(X,\mathcal B,m)$, and the operator is bounded. Now consider that, for any subset $Y\subset X$, $Y\in\mathcal B$, the function $$g_Y(x)=\frac{\chi_Y(x)}{(\text{meas(Y)})^{1/p}}$$ satisfies $$\|g_Y\|_{L^q}= \frac{1}{(\text{meas}(Y))^{1/p-1/q}}. $$ But then, for any $Y\subset X$, $Y\in\mathcal B$, we have $$\frac{1}{(\text{meas}(Y))^{1/p-1/q}}\leq \|G\|,$$ which means $$0<\frac{1}{\|G\|^{1/p-1/q}}\leq \text{meas}(Y).$$ Hence the result is proved.

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    $\begingroup$ A slightly better counterexample is given by $f_p(x)=x^{-1/p}$; this function belongs to $L^q(1,\infty)\setminus L^p(1,\infty)$ provided $q>p$. $\endgroup$ Commented Sep 20, 2011 at 12:23
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    $\begingroup$ In the second theorem (the first after the add), how do you show that the embedding $G\colon L^q\to L^p$ is bounded? $\endgroup$ Commented Sep 20, 2011 at 16:02
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    $\begingroup$ I'm sure you made a mistake in the proof of the last theorem. Now suppose $f$ is a simple function, i.e. $f(x)=\sum_{j=1}^n a_j\chi_{E_j}$, where $E_j$ are disjoint, the $L^q$ norm of $f$ should be $$\|f\|_{L^q}=\left(\sum_{j=1}^n a_j^q\mu(E_j)\right)^{1/q}$$. $\endgroup$
    – Xiang Yu
    Commented Oct 16, 2015 at 10:24
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    $\begingroup$ @juaninf Most probably, $\operatorname{meas}X=m(X)$ is the $m$-measure of $X$. $\endgroup$
    – 0xbadf00d
    Commented May 2, 2017 at 13:21
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    $\begingroup$ I think your proof of the first theorem after the ADD, as stated, is incorrect. You wrote in the proof: "By Hölder inequality, $$\|f\|_{L^p}\leq\text{meas }(X)^{1/r}\|f\|_{L^q}."$$ But, by applying Hölder inequality in this way, you are (implicitly) assuming that $\chi_X \in L^q$, which means $X$ has finite measure. However, the fact that $X$ has finite measure is exactly what you want to prove. $\endgroup$
    – Ramiro
    Commented Dec 12, 2020 at 17:44

In Rudin's book Real an complex analysis, we can find the following result, shown by Alfonso Villani:

Let $(X,\mathcal B,m)$ be a $\sigma$-finite measure space, where $m$ is a non-negative measure. Then the following conditions are equivalent:

  1. We have $L^p(X,\mathcal B,m)\supset L^q(X,\mathcal B,m)$ for some $p,q$ with $1\leqslant p<q<\infty$.
  2. $m(X)<\infty$.
  3. We have $L^p(X,\mathcal B,m)\supset L^q(X,\mathcal B,m)$ for all $p,q$ with $1\leqslant p<q<\infty$.

We only have to show that $1.\Rightarrow 2.$ and $2.\Rightarrow 3.$ since $3.\Rightarrow 1.$ is obvious.

$1.\Rightarrow 2.$: the inclusion $L^q(X,\mathcal B,m)\subset L^p(X,\mathcal B,m)$ is continuous. Indeed, let $\{f_n\}$ be a sequence in $L^q$ which converges to $f$ for the $L^q$ norm, and to $g$ for the $L^p$ norm. We extract a subsequence which converges almost everywhere to $f$ and $g$ (first extract a subsequence $\{f_{n_j}\}$ which converges to $f$ almost surely; this sequence still converges to $g$ for the $L^p$ norm; now extract from this sequence a subsequence which converges to $g$ almost surely), hence $f=g$, and by the closed graph theorem we get the conclusion since both $L^p$ and $L^q$ are Banach spaces.

Therefore, we can find $C>0$ such that $\lVert f\rVert_p\leqslant C\lVert f\rVert_q$. Since $X$ can be written as an increasing union of finite measure sets $A_n$, we get that $m(A_n)^{\frac 1p}\leqslant Cm(A_n)^{\frac 1q}$, hence $m(A_n)^{\frac{q-p} {pq}}\leqslant C$ and since $p\neq q$: $m(A_n)\leqslant C^{\frac{pq}{p-q}}$. Now we take the limit $n\to\infty$ to get $m(X)\leqslant C^{\frac{pq}{p-q}}$.

$2.\Rightarrow 3.$: let $1\leqslant p<q<\infty$ and $f\in L^q$. We put $E_n:= \left\{x\in X: \frac 1{n+1}\leqslant |f(x)|\lt\frac 1n\right\}$ for $n\in\mathbb N^*$. The sets $\{E_n\}$ are pairwise disjoint and by $2.$ we get $\displaystyle\sum_{n=1}^{\infty} m(E_n)<\infty$. The function $f$ is integrable because \begin{align*} \int_X |f|^pdm &=\int_{\{|f|\geq 1\}}|f|^pdm+\sum_{n=1}^{+\infty}\int _{E_n}|f|^pdm\\\ &\leqslant\int_X |f|^qdm+\sum_{n=1}^{+\infty}\frac 1{n^p}m(E_n)\\\ &\leqslant \int_X |f|^qdm+\sum_{n=1}^{+\infty}m(E_n)<\infty. \end{align*} Now we look at the case $q=+\infty$. If $m(E)<\infty$, since for each $f\in L^q$ we can find $C_f$ such that $|f|\leqslant C_f$ almost everywhere, we can see $f\in L^p$ for all $p$. Conversely, if $L^{\infty}\subset L^p$ for a finite $p$, then the function $f=1$ is in $L^p$ and we should have $m(E)<\infty$.

  • $\begingroup$ I don't see how you can just pick one subsequence in step 1 that converges to bot $f$ and $g$. It seems you can you just pick one for $f$ (or $g$), using Theorem 4.9 in Breviz, but how can you pick one that converges to both? $\endgroup$
    – csss
    Commented Apr 9, 2016 at 14:25
  • $\begingroup$ I have added details. $\endgroup$ Commented Apr 13, 2016 at 9:33
  • $\begingroup$ Shouldn't the $p,q$ verify $1\leqslant p<q \leq\infty$ at all steps instead of $1\leqslant p<q<\infty$? Bounded (essentially) function $f$ in a finite measure set implies $f \in L^p$ $\forall p \geq 1$. i.e. $L^{\infty} \subset L^p$. $\endgroup$
    – D1X
    Commented Jun 16, 2016 at 19:05

There is a easy way to show that. Suppose that $p<q$ and X a space measure finite. Take any $f\in L^q$. Then, the q-norm is finite. In this way, $$\int_X|f|^p = \int_{f(x)<1}|f|^p + \int_{f\geq1}|f|^p \leq\int1+\int_{f\geq1}|f|^q\leq\mu(X)+||f||_q^q<\infty$$ A counter example just take $$f(x)=\frac{1}{x}$$ for $x\in(0,\infty)$ and Lebesgue measure. Then $f$ belongs to $L^2$ (integral is 1) but not $L^1$.

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    $\begingroup$ You mean $x\in (1,\infty)$ here don't you? $\endgroup$
    – user161518
    Commented Jan 10, 2017 at 18:36
  • $\begingroup$ How did you get that $\int 1 \leq \mu(X)?$ $\endgroup$
    – homosapien
    Commented Jun 30, 2022 at 21:41
  • $\begingroup$ @Luiz ^^^^^^^^^^ $\endgroup$
    – homosapien
    Commented Jun 30, 2022 at 22:53
  • $\begingroup$ @user161518 can you explain how they got $\int 1 \leq \mu(X)$ $\endgroup$
    – homosapien
    Commented Jun 30, 2022 at 22:53
  • $\begingroup$ is it because its the integral split into two cases so the first part need bee less than the total measure of the space ? $\endgroup$
    – homosapien
    Commented Jun 30, 2022 at 22:54

This is to show that the restriction $1\leq p<q\leq\infty$ in the OP is not needed, and that the following result holds:

Theorem A: Suppose $(\Omega,\mathscr{F},\mu)$ is a $\sigma$-finite measure space. There exists $p,q$ with $0<p<q\leq\infty$ such that $L_q(\mu)\subset L_p(\mu)$ iff $\mu(X)<\infty$.

Sufficiency follows directly from Hölder's inequality. In fact, if $0<p<q\leq\infty$, $$\int_X|f|^p\,d\mu\leq \|f\|^p_q\big(\mu(X)\big)^{1-\tfrac{p}{q}}$$

Necessity can be shown using the simple and elegant argument described in Davide's posting. First some general observations about $L_p(\mu)$ spaces with $0<p<\infty$. To ease notation, we define $\|f\|_p=\Big(\int_X|f|^p\,d\mu\Big)^{1/p}$.

  1. The function $d_p:L_p(\mu)\times L_p(\mu)\rightarrow[0,\infty)$ given by $$d_p(f, g)=\Big(\int_X|f-g|^p\,d\mu\Big)^{\min(1,1/p)}=\|f-g\|_p^{\min(p,1)}$$ defines a complete translation invariant metric on $L_p(\mu)$, and so, $(L_p(\mu),d_p)$ is an $F$-space. (Of course when $p\geq1$, $L_p(\mu)$ is a Banach space with norm $\|f\|_p=d_p(0,f)$; for $0<p<1$, see this posting).
  2. The sets $B_p(0;r)=\{f\in L_p(\mu): d_p(0,f)<r\}$ form a local basis of neighborhoods for $(L_p(\mu),d_p)$.
  3. For any $r>0$, $B_p(0;r)=r^{\max(1,1/p)}B_p(0;1)$.

Now recall the general versions of the Closed Graph Theorem in the setting of $F$-spaces.

Theorem: Suppose $X$ and $Y$ are $F$-spaces. If $\Lambda:X\rightarrow Y$ is linear and its graph $G=\{(x,\Lambda x):x\in X\}$ is closed in $X\times Y$, then $\Lambda$ is continuous (and thus bounded since $X$ and $Y$ are metrizable).

This can be sen for example in Rudin, W., Functional Analysis, McGraw-Hill, second edition, 1991, pp. 51

Now we have all the ingredients to modify Davide's proof to the present setting. Suppose $L_q(\mu)\subset L_p(\mu)$ for some $0<p<q<\infty$. Consider the inclusion map $\iota: L_q(\mu)\rightarrow L_p(\mu)$. The continuity of $\iota$ follows from the Closed Graph Theorem just as explained by Davide: Suppose $(f_n:n \in\mathbb{N})\subset L_q(\mu)$ is a sequence which converges in $L_q(\mu)$ to some $f\in L_q(\mu)$, and which also converges in $L_p(\mu)$ to some $g\in L_p(\mu)$. Along a subsequence $f_{n'}$, $|f-f_{n'}|^q\rightarrow0$ $\mu$-a.s. Then, along a subsequence $f_{n''}$ of $f_{n'}$, $|f_{n''}-g|^p\rightarrow0$ $\mu$-a.s. It then follows that $f=g$ $\mu$-a.s. and so $\iota$ is bounded.

The key to extend the statement of the OP to the present setting is to understand how bounded sets in general $L_p(\mu)$ spaces look like.

Claim: For any $0<r\leq \infty$, $E\subset L_r(\mu)$ is bounded iff $$\sup\left\{f\in E: \|f\|_r\right\}<\infty$$

Proof: By definition (see Rudin, cit. op.) $E$ is bounded iff for any open neighborhood $U$ of $0$ in $L_p(\mu)$ there is $s>0$ such that $E\subset sU=\{sf: f\in U\}$. Notice that $$\|\lambda f\|^{\min(r,1)}_r=d_r(0,\lambda f)=\lambda^{\min(r,1)}\,d_r(0,f)=\lambda^{\min(r,1)}\|f\|^{\min(1,r)}_r$$ The conclusion of the claim follows immediately from observations (2) and (3), and by taking $U=B_r(0;1)$.

In particular, by observation (3) we have that any ball $\overline{B}_p(0;r)=\{f\in L_p(\mu):d_p(0,f)\leq r\}$ is bounded in $L_p(\mu)$.

By the continuity of $\iota$, $\overline{B}_q(0;1)$ is bounded in $L_p(\mu)$ since $\overline{B}_q(0;1)$ is bounded in $L_q(\mu)$. This means that $$c:=\sup_{f\in \overline{B}_q(0;1)}\|f\|_p<\infty$$

Notice that for any $f\in L_q(\mu)\setminus\{0\}$, $\frac{1}{\|f\|_q}f\in \overline{B}_q(0;1)$. Hence $$\|f\|_p\leq c\|f\|_q,\qquad f\in L_q(\mu)$$ Let $(A_n:n\in\mathbb{N})\subset\mathcal{F}$ such that $A_m\nearrow \Omega$. Then $$\big(\mu(A_n)\big)^{\tfrac{1}{p}-\tfrac{1}{q}}\leq c$$ whence we conclude that $\mu(X)\leq c^{\tfrac{pq}{q-p}}<\infty$.

Finally the case $0<p<q=\infty$ is simpler for if $L_\infty(\mu)\subset L_p(\mu)$, then as $f\equiv1\in L_\infty(\mu)$, $$\|f\|_p=\big(m(X)\big)^{1/p}<\infty$$


I'd like to provide a proof of a result mentioned in @uforoboa's answer, in which the proof seems questionable.

Theorem Let $1 \leq p < q < \infty$, and $(X, \it{M}, \mu)$ be a measure space. If $\exists c > 0$ such that for all $A \in \it{M}$ either $\mu(A) = 0$ or $\mu(A) \geq c$, then $L^p(X, \mu) \subset L^q(X, \mu)$.

Proof: Let $f$ be a function in $L^p(X, \mu)$. By Tchebyshev's inequality,$$\mu(E_t) \leq (\frac{\|f\|_p}{t})^t$$ where $E_t = \{x:|f(x)| > t\}$. Note when $t > \|f\|_p$, $\:(\frac{\|f\|_p}{t})^t \to 0$ as $t \to \infty$.
That is to say, $\exists T > \|f\|_p$ such that, $$\mu(E_T) < (\frac{\|f\|_p}{T})^T < c$$ Hence $$\mu(E_T) = 0$$ Then $$\|f\|_q^q = \int |f|^q = \int_{E_T}|f|^q + \int_{E^c_T}|f|^q = \int_{E_T^c}|f|^q$$ Since $|f| \leq T$ on $E^c_T$ and $p < q$, $$\|f\|_q^q = \int_{E_T^c}|f|^q = \int_{E_T^c}|f|^p|f|^{q-p} \leq T^{q-p}\int_{E_T^c}|f|^p \leq T^{q-p}\|f\|_p^p < \infty$$ By we which we conclude $f \in L^q(X, \mu)$ as desired.


Alternatively one can take into account this proposition.

$\textbf{Proposition. }$Let $(X, \mathcal{A}, \mu)$ a finite measure space. Let $f$ be $\mathcal{A}$-measurable function and let $E_{n} = \{x \in X: (n - 1) \leq \lvert f (x) \rvert < n\}$. Then $f \in L^{p}(X)$ if and only if $$\sum_{n = 1}^{\infty} n^{p} \mu(E_{n}) < + \infty.$$

Using this proposition, we have that for $1 \leq q < p$, if $f \in L^{p}(X)$ then $$ \sum_{n = 1}^{\infty} n^{q} \mu(E_{n}) \leq \sum_{n = 1}^{\infty} n^{p} \mu(E_{n}) < + \infty. $$ Hence $f \in L^{q}(X)$. So we have the inclusion $$ L_{p}(X) \subset L_{q}(X). $$


I would like to provide a proof restricted to probability space. Recall that Jensen's inequality states that if $X$ is a random variable and $c$ is a convex function: $$ \mathbb{E}(c(x)) \geq c(\mathbb{E}(x)) $$

Then let $1 \leq p < q < \infty$, and set $c(x):=x^{\frac{q}{p}}$, since $\frac{q}{p}$ is greater than 1, then $c$ is convex, so for any $X \in L^q(\Omega, \mathcal{F}, \mathbb{P})$ $$||x||_p = (\mathbb{E}(|x|^p))^{\frac{1}{p}} = (c(\mathbb{E}(|x|^p)))^{\frac{1}{q}} \leq (\mathbb{E}(c(|x|^p)))^{\frac{1}{q}} = (\mathbb{E}(|x|^q))^{\frac{1}{q}} = ||x||_q$$ Then $||x||_p$ is well defined, this implies $X \in L^p(\Omega, \mathcal{F}, \mathbb{P})$, so $$ L^q(\Omega, \mathcal{F}, \mathbb{P}) \subseteq L^p(\Omega, \mathcal{F}, \mathbb{P}) $$ However, I don't think this proof works for general finite measure space since Jensen's inequality only works for probability space. See the related answer

Does Jensen's inequality still hold in general finite measure space?


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