Worked Continuity Problems for Parameter Integrals

Continuity Requires a Local Dominator

A parameter-dependent integrand can converge pointwise while its integral behaves badly. To prove continuity of an integral, the key task is to find, around each parameter value of interest, one integrable function that bounds every integrand in that neighborhood. This tutorial applies the Dominated Convergence Theorem to powers on \((0,1)\), decaying exponentials on \((0,\infty)\), and a family containing both.

All integrals below are Lebesgue integrals with respect to Lebesgue measure \(m\) restricted to the indicated interval. The interval \((0,1)\) has finite measure; \((0,\infty)\) has infinite, but sigma-finite, measure. The integrands are nonnegative, finite-valued, measurable real functions. We use the Dominated Convergence Theorem in this form: if \(f_n\to f\) \(m\)-a.e. and \(|f_n|\le g\) \(m\)-a.e. for one integrable \(g\), then \(\int f_n\,dm\to\int f\,dm\).

We also use the power-integrability criteria from the earlier material on power and logarithmic integrability: \(\int_0^1 x^q\,dm\) is finite exactly when \(q>-1\), and, for every real \(q\) and \(c>0\), \(\int_1^\infty x^q e^{-cx}\,dm\) is finite. The second criterion reflects the fact that exponential decay dominates every fixed power at infinity.

Powers on the Unit Interval

Proof. Fix \(a_0>-1\). Choose \(\delta>0\) so that \(a_0-\delta>-1\). For any \(a\) satisfying \(|a-a_0|<\delta\), we have \(a>a_0-\delta\). Since \(0<x<1\), the function \(s\mapsto x^s\) is decreasing, and therefore \(0\le x^a\le x^{a_0-\delta}\). The proposed dominator is integrable, because \(a_0-\delta>-1\) and \(\int_0^1x^{a_0-\delta}\,dm=1/(a_0-\delta+1)<\infty\).

Now let \(a_n\to a_0\). For all sufficiently large \(n\), \(|a_n-a_0|<\delta\); thus the same integrable dominator bounds \(x^{a_n}\). For every \(x\in(0,1)\), continuity of the real exponential expression \(x^a=\exp(a\ln x)\) in \(a\) gives \(x^{a_n}\to x^{a_0}\). The Dominated Convergence Theorem on \(((0,1),\mathcal{L},m)\) gives \(P(a_n)\to P(a_0)\). The sequential criterion for continuity on the real line proves continuity at \(a_0\), and \(a_0\) was arbitrary. Finally, for \(a>-1\), integrating \(x^a\) gives \(P(a)=[x^{a+1}/(a+1)]_0^1=1/(a+1)\). If \(a=-1\), the integral of \(1/x\) diverges at zero; if \(a<-1\), the power-integrability criterion also gives divergence. Thus the stated range is exact. \(\square\)

Decaying Exponentials on the Half-Line

Proof. Fix \(\lambda_0>0\), and consider parameters \(\lambda\) with \(|\lambda-\lambda_0|<\lambda_0/2\). Then \(\lambda>\lambda_0/2\), so for every \(x>0\), \(0\le e^{-\lambda x}\le e^{-(\lambda_0/2)x}\). The dominator is integrable on \((0,\infty)\), with integral \(2/\lambda_0\). For any sequence \(\lambda_n\to\lambda_0\), all sufficiently large \(n\) lie in this neighborhood, and \(e^{-\lambda_nx}\to e^{-\lambda_0x}\) for every \(x>0\). Applying the Dominated Convergence Theorem gives \(Q(\lambda_n)\to Q(\lambda_0)\); the sequential criterion gives continuity. For \(\lambda>0\), direct integration yields \(\int_0^R e^{-\lambda x}\,dx=(1-e^{-\lambda R})/\lambda\). As \(R\to\infty\), \(e^{-\lambda R}\to0\), so \(Q(\lambda)=1/\lambda\). \(\square\)

Combining a Power with Exponential Decay

Proof. Fix \((a_0,\lambda_0)\) in the stated parameter set. Choose \(\delta>0\) such that \(a_0-\delta>-1\). Restrict parameters to \(|a-a_0|<\delta\) and \(|\lambda-\lambda_0|<\lambda_0/2\). On \(0<x<1\), \(a>a_0-\delta\) implies \(x^a\le x^{a_0-\delta}\), and \(e^{-\lambda x}\le1\). On \(x\ge1\), \(a<a_0+\delta\) implies \(x^a\le x^{a_0+\delta}\), while \(\lambda>\lambda_0/2\) implies \(e^{-\lambda x}\le e^{-(\lambda_0/2)x}\). Thus the family is bounded by the single function \(g(x)=x^{a_0-\delta}\) for \(0<x<1\) and \(g(x)=x^{a_0+\delta}e^{-(\lambda_0/2)x}\) for \(x\ge1\). Its first piece is integrable because \(a_0-\delta>-1\); its second is integrable by the power-exponential criterion. Therefore \(g\in L^1(m)\).

For any sequence \((a_n,\lambda_n)\to(a_0,\lambda_0)\), the parameters are eventually in the chosen neighborhood. At each \(x>0\), \(x^{a_n}e^{-\lambda_nx}\to x^{a_0}e^{-\lambda_0x}\). The same \(g\) dominates the sequence eventually, so the Dominated Convergence Theorem gives convergence of the integrals. The sequential criterion in \(\mathbb{R}^2\) proves continuity. The same bounds at any fixed admissible point also prove finiteness of the integral. \(\square\)

What Fails at a Boundary

The parameter restrictions are substantive. For powers, the endpoint \(a=-1\) is not part of the continuity domain: the integral is infinite there. A common incorrect domination argument is to claim that \(x^a\le1\) for negative \(a\) on \((0,1)\). For example, with \(a=-\tfrac12\) and \(x=\tfrac14\), \(x^a=(\tfrac14)^{-1/2}=2>1\). The inequality is reversed: negative powers grow near zero. The correct local dominator uses the smallest nearby exponent, which must remain greater than \(-1\).

For exponentials, the positive rate is essential for integrability on the half-line. At \(\lambda=0\), the integrand is \(1\), whose integral on \((0,\infty)\) is infinite. More generally, as \(\lambda\downarrow0\), \(Q(\lambda)=1/\lambda\) diverges. There cannot be an integrable dominator valid for rates in a neighborhood that includes zero. Pointwise convergence by itself does not repair this: \(e^{-\lambda_nx}\to1\) for each \(x>0\) when \(\lambda_n\downarrow0\), but the limiting function is not integrable on this measure space.

The practical procedure is to fix an interior parameter value first, choose a neighborhood that stays inside the valid range, and then bound every nearby integrand by one function whose integral is finite. DCT then proves continuity; it does not justify differentiation with respect to the parameter, which requires additional hypotheses.